' 


This  book  is  DUB  on  the  last  date  stamped  below 


OCT  31  1924 

;.«      ia^ 

AJOV  low 

*     47    XJ 

JAN  3 

til 

1927 

1  AUG  2  9  1938 

L*^-^' 

MAY  2  5 

192T 

WAR  ft      1940  ; 

Oo*»» 

.; 

<?r 

J\)L  6    t» 

-U/27 

JUL  1      1929  At 

^-    MAR  3    1945 

- 

r 

APR1  81949 

JUL  2  9 

1929 

AUG  2  7  1952 

R  3 

[930 

APR  2  0  1953 

f       MAY  4 
HAY  *8 

1931 
1931 

-JUL  101963    -- 

Dt^'|\  •  ri  tfi 
|\Hv  I/  LU'lJt 

IE  MAR  31 

^975 

JUN  8     1^31 

MAK    61 

975" 

^AN  22 

1931 

Dl^CHAQ 

nr  ,.n 

•     JAN  1  4 

^n 

JUNO 

19W 

JAN    3 

1980 

Form  L-9—  5m-5, 

'24 

UNIFIED 
MATHEMATICS 


BY 

LOUIS  C.  KARPINSKI,  PH.D. 

ASSOCIATE   PROFESSOR  OF    MATHEMATICS 
UNIVERSITY   OF   MICHIGAN 

HARRY  Y.  BENEDICT,  PH.D. 

PROFESSOR     OF     APPLIED     MATHEMATICS 
UNIVERSITY   OF   TEXAS 


JOHN  W.   CALHOUN,   M.A. 

ASSOCIATE    PROFESSOR    OF    PURE    MATHEMATICS 
UNIVERSITY    OF  TEXAS 


D.    C.    HEATH   &   CO.,   PUBLISHERS 
BOSTON  NEW  YORK  CHICAGO 


COPYRIGHT,  1918, 
BY  D.  C.  HEATH  &  Ca 

118 


/4 


PREFACE 

Tins  text  presents  a  course  in  elementary  mathematics  adapted  to 
the  needs  of  students  in  the  freshman  year  of  an  ordinary  college  or 
technical  school  course,  and  of  students  in  the  first  year  of  a  junior 
college.  The  material  of  the  text  includes  the  essential  and  vital 
features  of  the  work  commonly  covered  in  the  past  in  separate  courses 
in  college  algebra,  trigonometry,  and  analytical  geometry. 

The  fundamental  idea  of  the  development  is  to  emphasize  the  fact 
that  mathematics  cannot  be  artificially  divided  into  compartments 
with  separate  labels,  as  we  have  been  in  the  habit  of  doing,  and  to 
show  the  essential  unity  and  harmony  and  interplay  between  the  two 
great  fields  into  which  mathematics  may  properly  be  divided ;  viz., 
analysis  and  geometry. 

A  further  fundamental  feature  of  this  work  is  the  insistence  upon 
illustrations  drawn  from  fields  with  which  the  ordinary  student  has 
real  experience.  The  authors  believe  that  an  illustration  taken  from 
life  adds  to  the  cultural  value  of  the  course  in  mathematics  in  which 
this  illustration  is  discussed.  Mathematics  is  essentially  a  mental 
discipline,  but  it  is  also  a  powerful  tool  of  science,  playing  a  won- 
derful part  in  the  development  of  civilization.  Both  of  these  facts 
are  continually  emphasized  in  this  text  and  from  different  points  of 
approach. 

The  student  who  has  in  any  sense  mastered  the  material  which  is 
presented  will  at  the  same  time,  and  without  great  effort,  have 
acquired  a  real  appreciation  of  the  mathematical  problems  of  physics, 
of  engineering,  of  the  science  of  statistics,  and  of  science  in  general. 

A  distinctly  new  feature  of  the  work  is  the  introduction  of  series  of 
"  timing  exercises  "  in  types  of  problems  in  which  the  student  may  be 
expected  to  develop  an  almost  mechanical  ability.  The  time  which 
is  given  in  the  problems  is  wholly  tentative ;  it  is  hoped,  in  the 
interest  6f  definite  and  scientific  knowledge  concerning  what  may  be 
expected  of  a  freshman,  that  institutions  using  this  text  will  keep  a 
somewhat  detailed  record  of  the  time  actually  made  by  groups  of 
their  students.  The  authors  invite  the  cooperation  of  teachers  of  ele- 

iii 


iv  PREFACE 

mentary  college  mathematics  in  the  attempt  to  secure  this  valuable 
information.  The  authors  will  make  every  effort  to  put  information 
thus  secured  at  the  service  of  the  public  interested. 

In  general,  the  diagrams  are  carefully  drawn  on  paper  with  sub- 
divisions of  twentieths  of  an  inch.  It  is  expected  that  this  kind  of 
paper  will  be  used  as  far  as  possible  in  the  graphical  work,  as  students 
will  be  found  to  acquire  rapidly  the  ability  to  use  intelligently  this 
type  of  coordinate  paper.  Considerable  attention  should  be  paid  by 
the  teacher  to  the  intelligent  reading  and  interpretation  of  the  dia- 
grams which  appear  in  the  text,  as  the  student  will  in  this  way  gain 
power  to  handle  his  own  diagrams,  and  appreciation  of  the  vital 
importance  of  the  method.  The  photographic  illustrations  should 
also  be  used  in  a  somewhat  similar  manner. 

The  material  can  be  covered  without  systematic  omissions  in  a 
course  which  devotes  five  hours  per  week  for  one  year  to  the  study  of 
mathematics.  In  a  four-hour  course  there  are  certain  omissiims 
which  can  be  made  by  the  teacher  at  his  own  discretion ;  the  three 
chapters  on  solid  analytical  geometry  are  not  commonly  presented  in 
the  ordinary  four-hour  course ;  the  chapter  on  "  Poles  and  Polars  " 
may  also  be  omitted.  The  exercises  are  so  numerous  that  any  teacher 
can  make  a  selection,  which  can  be  varied,  if  desired,  in  succeeding 
years. 

No  attempt  has  been  made  to  introduce  the  terminology  of  the 
calculus  as  it  is  found  that  there  is  ample  material  in  the  more  ele- 
mentary field  which  should  be  covered  before  the  student  embarks 
upon  what  may  properly  be  called  higher  mathematics.  However, 
the  fundamental  idea  of  the  derivative  is  presented  and  utilized  with- 
out the  new  terminology. 

The  authors  are  greatly  indebted  to  a  large  number  of  their  col- 
leagues who  have  been  most  generous  in  furnishing  real  illustrations 
in  various  fields.  Professor  N.  H.  Williams  of  the  Department  of 
Physics  at  the  University  of  Michigan  has  given  very  pertinent  and 
valuable  comment  on  numerous  sections,  in  addition  to  furnishing 
the  beautiful  oscillograms  of  alternating  currents.  Professor  W.  J. 
Hussey  of  the  Detroit  Observatory  furnished  the  temperature  and  ba- 
rometer chart,  and  has  given  generously  of  his  time  in  the  discussion  of 
astronomical  problems  adapted  to  an  elementary  text.  Professors  J. 
J.  Cox,  H.  E.  Riggs,  A.  F.  Greiner,  H.  H.  Higbie,  J.  C.  Parker,  Leon 
J.  Makielski,  E.  M.  Bragg,  H.  TV.  King,  and  L.  M.  Gram  of  the  De- 
partment of  Engineering,  University  of  Michigan,  have  given  valu- 


PREFACE  V 

able  advice  and  suggestions.  The  diagram  illustrating  the  use  of  the 
ellipse  in  determining  the  proper  amounts  of  sand  and  gravel  to  use 
from  given  pits  to  obtain  the  best  results  was  furnished  by  Professor 
Cox.  To  Professor  Greiner  we  are  indebted  for  the  cut  of  the  six 
cylinders  of  an  automobile  engine,  and  for  criticising  the  piston- 
rod  motion.  To  Mr.  Makielski,  the  well-known  artist,  we  are 
indebted  for  the  drawing  of  a  box  which  is  reproduced.  Professor 
James  "W.  Glover  of  the  Actuarial  and  Statistical  Department,  Uni- 
versity of  Michigan,  has  read  and  corrected  the  material  relating  to 
his  field.  To  Professor  C.  L.  Meader  of  the  Department  of  Lin- 
guistics, and  to  Professors  Pillsbury  and  Shepard  of  the  Department 
of  Psychology,  University  of  Michigan,  we  are  indebted  for  the  tuning- 
fork  records  and  for  the  vowel  and  consonant  records.  To  Professor 
F.  G.  Novy  of  the  Hygienic  Laboratory  we  are  indebted  for  certain 
information  concerning  bacterial  growth.  Captain  Peter  Field,  Coast 
Artillery,  U.S.A.,  has  indicated  to  us  certain  simple  problems  con- 
nected with  artillery  work.  To  Mr.  H.  J.  Karpinski  we  owe  the 
photographs  of  the  Rialto  and  the  Colosseum.  To  the  Albert  Kahn 
Company  of  Detroit  we  are  indebted  for  information  concerning  de- 
tails of  the  Hill  Auditorium,  and  to  the  Tyrrell  Engineering  Company 
of  Detroit  for  permission  to  reproduce  a  number  of  photographs  of 
bridges.  We  render  to  these  gentlemen  and  to  our  colleagues  who 
have  been  generous  in  giving  time  and  thought  to  our  inquiries  our 
sincere  appreciation  for  their  friendly  cooperation.  In  every  field 
which  we  touch,  we  assume  full  responsibility  for  all  errors,  and  we 
shall  be  grateful  to  teachers  who  will  assist  in  removing  the  in- 
evitable blemishes  in  a  book  of  this  size  and  character. 

The  proof  has  been  carefully  read  by  Professor  E.  V.  Iluntington 
of  Harvard  University  and  by  Professor  C.  N.  Moore  of  the  University 
of  Cincinnati.  Many  blemishes  have  been  removed  and  many  impor- 
tant additions  and  changes  have  been  made  on  their  advice.  Pro- 
fessor J.  W.  Bradshaw  of  the  University  of  Michigan  and  Professor 
J.  D.  Bond  of  the  Texas  Agricultural  College  have  read  the  galley 
proof  and  given  numerous  and  excellent  suggestions.  Professor 
W.  W.  Beman  of  the  University  of  Michigan  has  read  the  page  proof 
and  has  made  numerous  vital  corrections  and  suggestions.  Professor 
J.  L.  Markley  has  given  advice  on  the  early  chapters.  To  all  of 
these  gentlemen  we  acknowledge  our  real  indebtedness. 

In  putting  the  work  through  the  press,  the  responsible  editorship 
has  been  placed  in  the  hands  of  Professor  Karpinski,  as  the  exigencies 


vi  PREFACE 

of  time  and  space  —  Texas  to  .Michigan  to  New  York  to  Boston  — 
would  have  delayed  the  book  for  a  full  year  with  a  divided  responsi- 
bility. Certain  chapters,  including  the  chapter  on  the  applications 
of  the  conic  sections,  the  chapters  on  the  sine  curve,  on  the  growth 
curve  and  on  complex  numbers,  the  treatment  of  solid  analytics, 
the  tables  and  most  of  the  problems,  are  due  entirely  to  Professor 
Karpinski. 

The  drawings  have  been  made  at  the  University  of  Michigan,  chiefly 
by  Mr.  E.  T.  Cranch,  an  engineer  now  in  the  service,  I'.S.A.  Most 
of  the  photographs  are  by  Miss  F.  J.  Dunbar  of  the  University  of 
Michigan  Lantern  Slide  Shop,  and  a  few  are  by  Mr.  G.  R.  Swain. 


CONTENTS 

CHAPTER  PAGE 

I.    NUMBERS  OF  ALGEBRA 1 

II.    APPLICATION  OF  ALGEBRA  TO  ARITHMETIC        .        .  18 

III.  EXPONENTS  AND  LOGARITHMS 40 

IV.  GRAPHICAL  REPRESENTATION  OF  FUNCTIONS      .        .  58 
V.    THE  LINEAR  AND  QUADRATIC  FUNCTIONS  OF  ONE 

VARIABLE 77 

VI.    STRAIGHT-LINE  AND  Two-PoiNT  FORMULAS       .        .  101 

VII.    TRIGONOMETRIC  FUNCTIONS 110 

VIII.    TABLES  AND  APPLICATIONS 139 

IX.    APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS   .        .  149 
X.    ARITHMETICAL  SERIES  AND  ARITHMETICAL  INTERPO- 
LATION       166 

XI.    GEOMETRICAL  SERIES  AND  APPLICATIONS  TO  ANNU- 
ITIES          176 

XII.    BINOMIAL  SERIES  AND  APPLICATIONS         .        .        .  193 

XIII.  RIGHT  TRIANGLES      .        .        .       ,        .        .        .206 

XIV.  THE  CIRCLE 220 

XV.    ADDITION  FORMULAS 237 

XVI.    TRIGONOMETRIC  FORMULAS  FOR  OBLIQUE  LINES        .  251 

XVII.    SOLUTION  OF  TRIANGLES    ......  266 

XVIII.    THE  ELLIPSE 281 

XIX.    THE  PARABOLA  .        .        .        .        .        .        .        .309 

XX.    THE  HYPERBOLA 320 

XXI.    TANGENTS  AND  NORMALS  TO  SECOND  DEGREE  CURVES  332 

XXII.    APPLICATIONS  OF  CONIC  SECTIONS     ....  346 

XXIII.    POLES,  POLARS,  AND  DIAMETERS       ....  363 

vii 


Vlll 


CONTENTS 


CHAPTER  PAGE 

XXIV.    ALGEBRAIC   TRANSFORMATIONS   AND   SUBSTITUTIONS  371 

XXV.    SOLUTION  OF  NUMERICAL  ALGEBRAIC  EQUATIONS     .  392 

XXVI.    WAVE  MOTION .        .        .407 

XXVII.    LAWS  OF  GROWTH 423 

XXVIII.    POLAR  COORDINATES 435 

XXIX.    COMPLEX  NUMBERS 439 

XXX.    SOLID   ANALYTIC    GEOMETRY:     POINTS   AND    LINES  452 
XXXI.    SOLID  ANALYTICS:    FIRST  DEGREE  EQUATIONS  AND 

EQUATIONS  IN  Two  VARIABLES      ....  465 

XXXII.    SOLID  ANALYTICS:  QUADRIC  SURFACES      .        .        .  475 

TABLES  495 


CHAPTER   I 
NUMBERS   OF  ALGEBRA 

-5         -4          -3         -2          -1  "          +1          +2         +3         4-4         46 

1.  Representation  of  points  on  a  line.  — With  any  given  unit 
of  length  and  a  fixed  point  of  reference,  called  the  origin,  the 
points  upon  a  given  line  are  located  by  numbers.  The  unit  of 
length  and  parts  thereof  are  laid  off  in  both  directions  from  the 
origin  to  locate  further  points.  To  each  point  corresponds  one 
number  (a  symbol)  and  only  one,  and  to  each  number  corre- 
sponds one  and  only  one  point.  We  call  this  a  one-to-one  corre- 
spondence. To  any  point  upon  the  line  of  reference  corresponds 
evidently  another  point  symmetrically  placed  with  respect  to 
the  origin.  This  symmetry  is  indicated  in  the  symbols  by 
using  the  same  set  of  symbols  twice,  distinguishing  by  two 
"  quality  "  signs  -f  and  — .  All  points  on  one  side  of  0  have  the 
+  sign  prefixed  to  the  symbols  designating  them,  while  the 
corresponding  points  on  the  other  side  take  the  same  symbols, 
prefixing  the  negative  sign.  Thus  +a  and  —  a  represent  sym- 
metrically placed  points  on  the  scalar  line.  The  line  of  refer- 
ence is  now  called  a  directed  line.  Of  two  numbers  represented 
by  points  on  this  line,  the  one  represented  by  that  one  of  the 
two  points  which  lies  to  the  right  hand  is  called  the  greater. 
Such  a  line  is  the  line  on  an  ordinary  thermometer ;  to  each 
number,  then,  there  corresponds  further  a  certain  temper- 
ature. Thousands  of  physical  and  material  interpretations  of 

1 


2  UNIFIED  MATHEMATICS 

the  points  upon  such  a  line  and  the  corresponding  numbers 
are  possible. 

This  scalar  line,  as  the  above  is  termed,  is  not  necessarily  a 
straight  line.  Thus  the  equator  is  a  scalar  line  as  it  is  repre- 
sented upon  any  globe,  with  the  zero  at  the  intersection  with 
the  meridian  of  Greenwich,  and  distances  given  in  degrees, 
each  representing  ^i^  part  of  the  equatorial  circumference  of 
the  earth  ;  -f  and  —  are  represented  on  this  line  by  E.  and  \V. 

PROBLEMS 

1.  Interpret  a  scalar  line  as  representing  distance  upon  the 
main  line  of   the   Michigan   Central   Railroad  from  Detroit, 
east  and  west. 

2.  What  is  the  significance  of  points  to  the  left  of  the  origin 
when  the  line  represents  your  bank  account  ? 

3.  Interpret  the  line  as  representing  weights. 

4.  Interpret  the  line  as  the  prime  meridian.     What  length 
is  represented  by  1°  (circumference  of  earth  is  25,000  miles)  ? 

5.  Interpret  the  scale  as  representing  percentage  of  fat  in 
foods. 

6.  Represent  the   Fahrenheit  scale  on  one  side  of   such  a 
line  and  the  Centigrade  upon  the  other,  making  the  zero  and 
the  100°  of  the  Centigrade  scale  fall  upon  the  32°  and  212°  of 
the  Fahrenheit ;   note  that  for  a  convenient  total  length  20° 
Fahrenheit  may  be  taken  as  corresponding  to  1  centimeter  or 
to  one  half  an  inch. 

2.  Real  numbers :  positive  integers.  —  The  symbols  repre- 
senting the  points  upon  a  line,  as  above,  are  called  real  num- 
bers. Elementary  algebra  is  largely  a  study  of  such  numbers, 
combined  according  to  certain  rules.  The  rules  of  the  game 
of  algebra,  as  we  may  term  it,  can  be  studied  entirely  apart 
from  any  physical  application,  but  the  study  is  of  funda- 
mental importance  because  of  the  part  which  algebraic  num- 
bers play  in  the  sciences.  However,  a  knowledge  of  the  laws 


NUMBERS  OF  ALGEBRA  3 

of  algebra,  apart  from  the  applications,  is  necessary  to  enable 
one  to  apply  the  numbers  effectively  to  physical  problems. 

The  real  numbers  are  sub-divided  into  positive  and  negative 
numbers  ;  another  classification  is  into  rational  and  irrational 
numbers,  the  rational  numbers  being  further  sub-divided  into 
integers  and  fractions.  Numbers  are  represented  by  the  letters 
a,  b,  c,  -  x,  y,  z,  etc. 

Integers  were  undoubtedly  conceived  long  before  man  began 
to  write.  The  idea  of  an  integer  involves  the  notion  of  a 
group  of  individual  objects,  and  of  one-to-one  correspondence. 
The  idea  or  notion  which  is  common  to  all  groups  of  objects 
which  can  be  placed  in  one-to-one  correspondence  with  the 
objects  of  a  given  group  is  called  the  number  of  the  given 
group  of  objects.  Thus  the  pennies  OOOOO  can  be  placed 
in  one-to-one  correspondence  with  some  segments  of  our  line, 
or  with  the  group  of  symbols  which  correspond  to  these 

OOOOO 


segments,  or  with  the  individuals  of  any  one  of  infinitely 
many  other  groups,  of  number  Jive,  which  have  the  one  com- 
mon property  that  they  can  be  placed  in  one-to-one  correspond- 
ence with  each  other.  The  definition  is  recent ;  the  idea  is 
old.  One-to-one  correspondence  appears  frequently  in  physical 
problems,  as  in  the  one-to-one  correspondence  between  degrees 
Centigrade  and  degrees  Fahrenheit  above. 

Integers  can  be  used  to  represent  segments  of  our  line  of 
reference,  from  0  as  reference  point,  with  some  length  as 
unit  of  measure  (or  as  individual  of  the  group).  The  ex- 
tremity farthest  from  0  is  marked  with  the  integer  corre- 
sponding to  the  number  of  the  segments  between  that  point 
and  0.  Evidently  certain  groups  of  segments  include  as  sub- 
groups other  groups  of  segments.  The  number  of  the  includ- 
ing group  is  called  greater  than  the  number  of  any  included 
group ;  the  included  group  is  smaller,  and  its  number  is  less 
than  the  number  of  the  including  group.  Thus  the  group 


4  UNIFIED  MATHEMATICS 

called  eight,  8,  has   the  smaller   sub-groups,  1,  2,  3,  4,  5,  6, 
and  7. 

1       1       I       2       1       3       I       4~~l       5       I       6       I       7       I       8       1 

3.  Positive  integers ;  fundamental  laws,  definitions,  assumptions, 
and  theorems.  —  Given  two  positive  integers,  a  and  b,  the  single 
group  composed  of  the  individuals  from  two  distinct  groups 
of  objects,  represented  by  a  and  b  respectively,  is  represented 
by  another  number  x,  the  sum  of  a  and  6,  which  latter  we  terra 
summands.  The  process  of  finding  such  a  number  is  called 
addition,  and  is  indicated  by  writing  the  sign  +  between  the 
two  given  numbers  a  and  b.  By  the  sum  of  three  numbers  is 
meant  the  number  obtained  by  adding  the  third  to  the  sum  of 
the  first  two,  and  similarly  for  more  numbers  than  three.  The 
following  are  assumptions  and  theorems  concerning  integers. 

I.  x  =  a  +  6,  given  two  integers,  the  sum  exists. 

II.  a  +  6  =  b  +  a,  addition  is  commutative,  i.e.  the  order  of 
addition  is  immaterial. 

III.  a  +  b  +  c  =  (a  +  6)  +  c  =  a  +  (b  +  c);  the  associative  law 
for  addition. 

IV.  x  +  b  =  a.     Given  the  sum  a  and  one  of  the  summands, 
the  other  sumniand  exists :  x  is  the  number  which  added  to  b 
gives  a.     This  defines  the  operation  of  subtraction,  which  is 
represented  by  the  sign  — ,  to  be  placed  between  the  sum  and 
the  given  summand,  as  in  a  —  b  =  x. 

By  definition,  (a  —  6)  +  b  =  a,  and  for  the  present  this  has 
meaning  only  when  b  is  less  than  a ;  a  is  termed  minuend,  b  is 
termed  subtrahend,  and  a  —  b  is  the  remainder. 

Thus,  given  x  +  2  =  7,  x  is  evidently  5,  as  one  remembers  that  in  the 
operation  of  addition  5  added  to  2  gives  7.  Given  x  +  2  =  2,  or  «  +  2  =  1, 
we  have,  at  this  stage  of  development,  no  number  x  which  satisfies  the 
given  condition. 

V.  If  a  +  c  =  6  +  c,  a  =  6,  and  conversely. 

The  converse  is  equivalent  to  the  axiom,  if  equals  be  added 
to  equals  the  sums  are  equal. 


NUMBERS  OF  ALGEBRA  5 

VI.  x  =  a  •  b.     Suppose  that  each  individual  of  a  group  of 
a  objects  consists  of  b  individuals  of  another  type,  e.g.  4  rows, 
each  of  7  dots,  then  the  single  group 

consisting  of  all  the  second  type   of 

individuals  involved  is  called  the  prod- 

uct  of  a  and  b.   The  operation  is  called 

multiplication  and   is   represented  by 
the  sign  x  between  a  and  6,  or  by  a 

period  (slightly  elevated)  between  a  and  b,  or  by  simple  juxta- 
position of  -the  two  numbers,  a  and  6,  called  factors ;  a  is 
termed  multiplier,  and  b  is  multiplicand. 

VII.  a  •  b  =  b  •  a,  the  commutative  law  for  multiplication, 
evident  from  the  figure. 

VIII.  a  •  b  •  c  =  (a  •  6)  •  c  =  a  (b  •  c),,  the  associative  law  for 
multiplication. 

IX.  a(b  +  c)  =  ab  +  ac,  multiplication  is  distributive   with 
respect  to  addition.     This  corresponds  precisely  to  our  ordi- 
nary method  of  multiplication. 

X.  b  -  x  =  a.     Given  the  product  a,  and  one  factor  b,  x  is  de- 
fined by  this  relation  as  the  number  which  multiplied  by  b  gives 
a.     This  operation  is  limited  when  dealing  with  integers  to  num- 
bers a  and  b,  which  are  so  related  that  a  is  one  of  the  products 
obtained  by  multiplying  b  by  an  integer. 

The  process  of  finding  x  is  called  division,  and  is  represented 

by  the  sign  -*- ,  or  by  placing  a  over  6,  b  •  -  =  a ;  b  is  termed 

the  divisor,  a  is  the  dividend,  and  x  is  termed  the  quotient. 

These  laws  concerning  positive  integers  constitute  simply  a 
restatement  of  facts  with  which  the  student  is  familiar.  The 
four  fundamental  operations  to  this  point  have  been  confined 
to  the  field  of  positive  integers ;  evidently  the  operation  of 
division  when  6  is  the  divisor  applies  only  to  those  positive 
integers  which  are  multiples  of  b.  Similarly  the  operation  of 
subtraction  of  b  from  a  is  limited  to  integers  so  related  that 


6  UNIFIED  MATHEMATICS 

a>&.  We  extend  our  field  of  numbers  by  removing  these 
limitations.  Thus  if  you  wish  to  have  a  number  x  which 
multiplied  by  8  gives  5  you  do  not  find  it  among  the  positive 
integers  ;  you  may  then  decide  to  create  such  a  number,  calling 
it  £,  the  two  symbols  indicating  the  definition  and  genesis  of 
the  new  number.  Such  extensions  of  the  number  field  are 
briefly  indicated  in  the  next  section. 

4.  Rational  numbers  ;  zero,  fractions,  and  negative  integers.  — 
These  fundamental  equalities  and  definitions  from  I  to  X  are 
now  extended  by  removing  all  limitations  (except  one,  as  noted 
below)  upon  the  numbers,  a,  b,  c,  and  x.  Note  that  only  the 
operations  of  addition,  subtraction,  multiplication,  and  division 
are  included  at  this  point. 

Extension  of  IV.  x  +  6  =  a,  when  b  =  a  defines  zero, 
written  0.  By  definition  then,  0  +  a  =  a.  x  +  b  =  0,  defines 
a  negative  number  which  is  written  —  b.  The  negative  here 
is  a  sign  of  quality ;  by  definition  —  6  is  the  result  of  subtract- 
ing 6  from  0,  and  —  b  -f-  b  =  0. 

x  -f  b  =  a,  when  b  >  a,  defines  the  negative  number  a  —  b, 
which  is  the  negative  of  b  —  a. 

Subtracting  a  negative  number  can  now  be  shown  to  be 
equivalent  to  adding  a  positive  number,  and  similarly  the  other 
rules  of  elementary  algebra  relating  to  the  addition  and  sul> 
traction  of  positive  and  negative  quantities.  That  the  product 
of  two  numbers  with  like  signs  is  positive  and  the  product  of 
two  numbers  with  unlike  signs  is  negative  follows  from  the 
above  development. 

A  negative  number  —  a  is  placed  in  our  line  of  reference 
symmetrically  to  the  corresponding  positive  number  a,  with 
respect  to  the  origin ;  of  two  negatives,  the  one  toward  the 
right  is  called  the  greater. 

Extension  of  X.     b  •  -  =  a,  for  all  values  of  b  except  0. 
b 

This  extension  of  -  to  mean  a  number  which  multiplied  by 


NUMBERS  OF  ALGEBRA  7 

b  gives  a  introduces  new  numbers  of  the  type  - ,  rational  frac- 
tions in  which  a  and  b  are  positive  or  negative  integers. 
Integers  are  included  in  this  definition  if  &  is  a  factor  of  a  or 
if  b  is  equal  to  one.  Division  by  0  is  explicitly  excluded. 

A  rational  number  is  any  number  which  can  be  expressed  as 
the  quotient  of  two  integers,  the  denominator  not  to  be  zero. 

All  of  our  rules  for  operating  with  fractions  follow  from  the 
definition  of  -  and  from  the  preceding  development.     Thus 
a  a      —  a 


-b 


Further,  by  definition, 


ft  f* 

->-,  when  both  are  positive  if  ad  >  be ; 
b      d 

fl  (*•  s 

-  =  — ,  when  both  are  positive  if  ad  =  be ;  and 
6     d 

-  <  - ,  when  both  are  positive  if  ad  <  be. 
b      d 

Positive  fractions  can  thus  be  arranged  in  a  determined  order 
upon  our  line  of  reference  ;  the  value  of  the  fraction  determines 

the  position  and   a   graphical  method  of  locating  -   on  the 

b 

scalar  line  is  indicated  in  the  next  section ;  negative  fractions 
are  placed  symmetrically  to  the  corresponding  positive  frac- 
tions, with  respect  to  the  origin. 

Of  any  two  rational  numbers  a  and  b,  a  is  greater  than  b 
(a  >  6)  if  a  —  6  is  positive ;  for  when  a  —  b  is  positive,  a  posi- 
tive length  must  be  added  to  b  to  give  a,  and  consequently  a 
must  lie  to  the  right  of  b.  If  on  the  line  of  reference  two 
points  ajx  and  x2  are  taken  (fixed  points),  x2  —  xl  gives  the  dis- 
tance from  the  first  point  to  the  second ;  this  expression  is 
positive  if  x2>  a?l7  and  negative  if  a?2<a^. 

That  there  is  a  distinction  between  +  and  —  used  as  signs 
of  operation,  as  with  positive  integers  in  the  preceding  section, 


8  UNIFIED  MATHEMATICS 

and  +  and  —  used  as  quality  signs  is  apparent.  Thus  —  6 
may  indicate  that  6  is  to  be  subtracted  from  some  preceding 
number,  or  —  b  may  indicate  that  the  distance  b  is  taken  on 
the  negative  side  of  the  origin.  The  fact  that  a +(—6),  the 
addition  to  a  of  negative  b,  gives  the  same  result  as  subtract- 
^  ing  b  from  a,  or  a  —  b,  is  readily  shown  by  the  graphical 
method  of  section  9  below.  This  type  of  relationship  obviates 
any  need  for  careful  distinction  between  the  two  possible  mean- 
ings of  these  signs  and  makes  separate  symbols  not  necessary. 
When  no  sign  is  used  with  a  number  symbol  the  +  sign  is 
understood. 

EXERCISES 

3  3 

1.  Explain  the  distinction  between  — —  and ;  between 

7  7 

3         ,  -3 
and  • 

—  7  7 

5  a 

2.  Write  -        -   in  the  three  forms  corresponding  to > 

3  —  x  b 

a        T     a 

, and  

6'          -b 

3.  Is  —  3  >  —  2  ?     Which  is  greater,  0  or  —  3  ?     Explain. 

4.  What  is  the  difference  between  4  and  —3?   4  and  3? 
4  and  11  ? 

5.  What  fundamental  law  is  assumed  in  the  common  process 
of  multiplication,  e.  g.  as  in  325  by  239  and  also  x—7  by  x—2? 
Is  there  a  corresponding  assumption  in  division  ? 

6.  Which  is  greater,  4-  or  —  f  ?     Is  ±1  greater  or  less  than 
f|  ?     Explain. 

7.  What  is  the  product  of  0  by  7 ;  by  -  8 ;  by  3 ;  by  ^?     If 
a  product  is  zero,  what  limitation  is  imposed  upon  the  factors  ? 

5.    Representation  of  a  rational  number,  —     On  cross-section 

6 

paper  any  rational  fraction  can  be  represented,  using  ruler  and 
compass.     Using  5  divisions  to  represent  unity,  each  division 


NUMBERS  OF  ALGEBRA 


9 


represents  4-  of  a  unit.  To  represent  ±f-,  one  measures  off  13 

units,  OA,  on  the  line  of   reference,  and  7  units,  OB,  on  a 

second  line  through  the  origin  (for  convenience,  use  cross-sec- 

-r-B-T, 


3     -'"4 


_ 


7.-    12  : 


Graphical  division 
\  to  $  of  13  represented  on  horizontal  line  of  reference. 

tion  paper).  Connect  the  ends,  AB,  and  through  the  point  U, 
one  unit  from  0  on  the  second  line,  draw  a  line  parallel  to  AB. 
The  intersection  point  on  the  reference  line  represents  the 

fraction  -aT3-.     Similarly  any  fraction  ^   can   be   represented. 

The  series  of  parallels  to  AB  through  the  first  7  unit  points 
on  the  vertical  axis  will  cut  off  (plane  geometry  theorem)  7 
equal  parts  of  13  on  the  horizontal  axis. 


Graphical  division 

to  -y-  expressed  decimally. 


10 


UNIFIED  MATHEMATICS 


On  cross-section  paper  a  somewhat  better  method  of  indicat- 
ing  any   quotient    -  is  to  move  out  on  the  line  of  reference  6 
b 

units  and  up  1  unit ;  connecting  this  point  with  the  origin  0 
gives  a  straight  line  which  can  be  used  to  read  the  desired 
quotients.  Thus,  since  OB  =  7  units,  EC  —  1,  and  OA  =  13, 

AP 

it  follows  that      —  = 


BC      OB' 


=      ,  whence  AP  equals       . 


8.5 


To  obtain  -^-,  you  find  the  point  8.5  units  from  0,  and  the 
7 

vertical  distance  to  the  oblique  line  represents  -^-,  or  1.2. 

7 


6.    Irrational  numbers.       s?  =  2  is  a  simple  and  familiar  illus- 
tration of  a  relation  which  is  not  satisfied  by  any  rational 

number,  -,  with  a  and  b  integers ;  geometrically,  the  diagonal 

of  a  square  with  side  unity 
is  not  represented  by  any 
rational  number.  If  you  wish 
the  length  of  this  diagonal 
for  any  practical  purpose,  you 
use  i£,  or  -u(3>  Or  \fo,  or  if, 
or  liii  or  1^42.  The  car. 
penter  uses  1  foot  5  inches, 
or  1^  feet,  in  the  diagonal 


for  every  foot  of  side,  with 
an  error  of  ^  of  one  per 
cent.  The  series  of  rational 
,  which  can  be  indefinitely  ex- 
tended always  increasing,  and  the  series,  always  decreasing, 
2>  TT>  TF! >  TW¥'  TOOOO>  w^h  a  constantly  decreasing  difference 
of  limit  0  between  corresponding  terms,  together  define  the 
irrational  number  called  the  square  root  of  2.  No  rational 

number  satisfies  the  relation ;   no  number  -  is  at  the  end  of 

b 
either  series,  but  either  series  determines  a  definite  point  on 


.2      .4      .6      .8       1      1.2  1.4 
Graphical  representation  of  V2 

numbers  1,  14,  | 


NUMBERS  OF  ALGEBRA 


11 


our  line,  and  algebraically  defines  our  number,  which  we  will 
call  the  square  root  of  2. 

Proof  of  the  irrationality  of  V2. —  Assume  that  V2  =->  a 
rational  fraction  in  lowest  terms,  with  p  and  q  integers.  Both 
p  and  q  cannot  be  even  numbers,  either  p  is  odd  or  q  is  odd. 
If  p  is  odd,  squaring  and  clearing  of  fractions,  2  3*  =  pz ;  but 
p  is  odd  and  you  have  an  even  number  equal  to  an  odd 
number.  Hence  p  cannot  be  an  odd  number.  Kow  assume 
that  p  is  even  and  that  q  is  odd,  and  further  let  p  =  2  ra. 

Then  2  q-  =  4  m*,  q9  =  2  m*,  and  again  we  have  an  odd  num- 

p 
ber  equal  to  an  even  number.     Our  assumption  that  V  2  =  - 

leads  to  an  absurdity,  that  an  odd  number  equals  an  even 
number. 

Describe  about  the  origin  with  a  radius  10  a  circle,  and 
using  a  protractor  measure  an  angle  of  20  degrees.  The 
length  of  the  perpendicular  and 
the  part  cut  off  by  the  perpen- 
dicular from  the  end  of  this 
line  are  definite  and  precise 
points  which  can  be  computed 
to  any  degree  of  accuracy  de- 
sired. No  rational  number  rep- 
resents these  lengths,  which  are 
trigonometric  irrationalities.  A 
series  of  constantly  increasing 
rational  numbers  can  be  found, 
such  that  there  is  no  greatest 
of  the  series,  to  represent  lines 
which  are  always  shorter  than  the  given  line ;  and  another 
series  of  terms  constantly  decreasing,  but  approaching  to  the 
terms  of  the  first  series,  can  be  found.  No  largest  number 
can  be  found  in  the  first  series  and  no  smallest  in  the  second ; 
both  sequences  together  define,  we  may  say,  an  irrational 
number. 


A  trigonometric  irrational 


12  UNIFIED  MATHEMATICS 

EXERCISES 

1.  Write  6  terms  of  the  decreasing  series  defining  V2;  V3; 
V5. 

2.  How  is  the  series  for  defining  the  length  of  the  circum- 
ference of  a  circle  obtained  ?     What  assumption  is  made  ? 

3.  Find  the  series  defining  the  square  root  of  3673  by  suc- 
cessive approximations. 

4.  Inscribe  a  square   and    an   equilateral   triangle   in   the 
circle  of  radius  10 ;  find  the  sides. 

7.  Constants  and  variables.  —  Every  fixed  point  on  the  line 
of  reference  is  at  a  fixed  distance  from  the  point  of  reference, 
0 ;  the  distance  is  constant.     We  can  think  of  a  point  as  mov- 
ing on  the  line  OX  in  either  direction.     The  distance  from  0 
then  varies  and  we  speak  of  the  distance  as  a  variable.     Thus, 
also,  the  price  of  wheat  during  a  term  of  years  or  in  different 
parts  of  the  world  is  a  variable  ;  the  weight  of  an  animal  at 
different  ages  is  a  variable.    We  think  of  the  variable  quantity 
as  taking  a  series  of  values  under  diverse  conditions.     We  can 
represent  the  variable  distance  of  a  point  from  0  on  our  line 
by  the  single  letter  cc,  which  may  then  in  the  various  possible 
positions  on  the  line  be  thought  of  as  positive,  or  negative,  or 
zero,  as  rational  or  irrational.     This  letter  x  represents  then  a 
variable  quantity  and  is  essentially  a  number,  subject  to  all 
the  operations  on  algebraic  quantities  as  noted  above.     In  gen- 
eral, we  designate  the  variable  point  by  the  single  letter  P, 
the  distance  from  0  is  OP,  of  which  the  length  and  direction 
from  0  is  indicated  by  the  number  or  variable  x.     A  point  on 
the  line  XfX  is  represented  by  a  single  letter  x,  called  the 
abscissa  of  the  point.     The  fixed  points  on  the  line  are  fre- 
quently represented  by  the  letters  a,  b,  c,  d,  •••  or  by  x1}  xz, 
£3,  •«,  each  of  which  may  represent  any  point  upon  the  line. 

8.  Historical   note. — Modern   algebra  with   the   systematic 
employment  of  literal  coefficients,  letters  to  represent  general 
constants,  was  introduced  by  the  great  French  mathematician 


NUMBERS  OF  ALGEBRA 


13 


and  statesman,  Francois  Viete  (1540-1603) ;  Viete  used  the 
consonants  to  represent  known  quantities  and  vowels  to  re- 
present unknowns,  using  capitals  for  both.  The  use  of  the 
symbols  of  operation  in  equations  dates  also  from  about  the 
same  time ;  our  equality  sign  was  introduced  by  Robert 
Recorde,  an  English  physician  and  mathematician,  whose 
"  Whetstone  of  Witte,"  1557,  is  the  first  treatise  in  the  Eng- 
lish language  on  algebra.  The  -f  and  —  symbols  are  due  to 
a  German,  Widrnann,  and  date  from  1489. 

9.  Geometrical  equivalents  of  the  four  fundamental  operations, 
a.  Addition.  —  The  operation  of  addition  of  xl  and  x2  is  repre- 
sented graphically  by  placing  the  length  OP2  upon  the  line 


1:17 


OP3  = 


+  PjP3  =  OPi  +  OP2  =  xi  +  xa. 


from  the  point  PI  in  the  direction  of  OP2.  Physically, 
addition  is  the  result  in  general  of  two  different  causes. 
Thus  a  weight  of  3  pounds  -f  a  weight  of  5  pounds  ;  a  vertical 
velocity  due  to  the  action  of  gravity  (on.  a  falling  body)  +  a 
vertical  velocity  due  to  some  other  force ;  a  transportation 
(translation)  from  one  point  to  another  +  another  translation 
in  the  same  direction ;  two  successive  rotations  of  a  wheel 
about  its  axis ;  these  are  familiar  examples  of  addition. 

6.  Subtraction.  OPi  +  PiP2  =  OP2 ;  P^P2  =  OP2  -  OP^ 
Whatever  the  relative  positions  on  the  line  OX  of  Pr  and  P2, 
with  respect  to  the  position  of  0,  OPt  +  P^P2  =  OP2,  all  of 
these  representing  directed  line  segments.  In  words,  the 
equality  PiP2  =  OP2  —  OPi  states  that  the  distance  from  any 
point  PI  on  a  directed  line  to  a  second  point  on  the  line  is 
given  by  the  abscissa  of  the  second  point  minus  the  abscissa  of 


14 


UNIFIED  MATHEMATICS 


the  first,  with  respect  to  any  third  point  O,  on  the  line,  as 
origin. 

If  we  represent  the  distance  from  Pl  to  P2  by  the  letter  d, 
we  have  Xi  +  d  =  x,2,  or  d  =  x-^  —  xt.  Subtraction  is  represented 
by  the  distance  from  the  first  point  to  the  second  point,  which 


OPi  +  PjP2  =  OP2. 
Fundamental  property  of  any  three  points  on  a  directed  line 

equals  OP2  —  OPi-  Since  in  physical  problems  the  distance 
represents  the  rise  or  fall  in  numerical  value  of  physical  values, 
graphical  subtraction  is  more  frequently  noted  than  addition. 
We  say  the  temperature  has  risen  10  degrees  or  fallen  10 
degrees,  having  in  mind  the  original  and  the  final  reading ; 
the  iron  bar  has  expanded  an  inch,  has  a  precise  and  valuable 
physical  meaning  when  the  initial  and  the  final  length  of  the 
bar  are  known.  The  operation  involved  in  measuring  the 
expansion  is  strictly  subtraction. 

c.  Multiplication  and  division.  —  Graphical  multiplication 
and  division  upon  cross-section  paper,  involving  theorems  con- 
cerning similar  triangles,  lends  itself  to  numerical  work  where 
great  accuracy  is  not  necessary.  Some  convenient  length  is 
taken  to  represent  1  unit  on  the  horizontal  scale  and  a  con- 
venient length  as  unit  on  the  vertical  scale ;  for  three-place 
accuracy  a  sheet  of  cross-section  paper  considerably  larger 
than  the  page  of  this  book  would  be  necessary.  If  multiples 
and  quotients  of  27  (2.7  or  270  or  2700  ...)  are  desired, 
take  the  unit  on  the  horizontal  scale  so  as  to  use  8  to  10 
inches  to  represent  the  numbers  up  to  10 ;  on  a  vertical  line 
above  the  10  on  the  horizontal  scale  represent  27.  On  our 


NUMBERS  OF  ALGEBRA 


15 


diagram  below  we  have  used  1  horizontal  unit  to  repre- 
sent 10  on  the  vertical  scale;  greater  accuracy  would  be 
attained  by  using  a  longer  length  as  unit  on  the  vertical  scale. 
The  tenths  of  27  are  represented  by  the  vertical  lengths  from 
the  points  1,  2,  3,  4,  ...,  9  on  the  horizontal  scale  to  the  line 
joining  the  intersection  of  the  two  scales  to  the  point  27  units 
above  the  point  10  on  the  horizontal  scale.  To  multiply  2.7 
by  3.6  you  find  the  vertical  length  above  the  point  3.6  on  the 


Graphical  multiplication 
Tenths  and  hundredths  of  27  represented  by  the  vertical  lines. 

horizontal  scale  and  this  length  is  9.7 ;  for  27  x  36  you  inter- 
pret this  as  970  and  similarly  for  .27  x  .36  you  interpret  this 
as  .097.  Evidently  MP :  3.6  =  AG  :  10,  or  MP :  3.6  =  27  : 10, 
whence  MP  =  3.6  x  2.7.  In  division  the  length  MP  corre- 
sponds to  the  dividend  and  27  to  the  divisor ;  the  quotient  is 
then  given  as  the  horizontal  distance  corresponding  to  the 
given  dividend  (MP).  Thus  20  -f-  27  is  obtained  graphically 
by  finding  the  horizontal  distance  at  which-  the  line  20  units 
above  our  horizontal  scale  cuts  the  given  oblique  line;  this 
appears  to  be  at  7.4  units  and  must  be  interpreted  as  .74; 
the  position  of  the  decimal  point  must  be  fixed  by  the  com- 
puter. The  explanation  would  be  a  trifle  easier  if  the  line  of 
length  27  were  placed  directly  above  the  unit  distance,  but  a 
slight  error  in  the  position  of  the  upper  point  would  produce 
large  errors  at  the  right  end  of  the  line ;  a  line  to  be  used  as 
indicated  must  be  kept  on  the  paper  and  it  must  be  drawn 
between  points  which  are  not  close  together. 


16  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  Represent  the  square  root  of  2  geometrically,  taking  10 
quarter-inches  as  1.    Represent  the  square  root  of  3  on  coordi- 
nate paper,  using  the  same  scale.     Represent  the  square  root 
of  5,  6,  7,  and  8. 

2.  On  coordinate  paper  take  a  horizontal  length  of  10  half- 
inches  and  at  the  right  extremity  draw  a  vertical  line  7  half- 
inches  long.     Draw  the  line  connecting  the  extremity  of  the 
vertical  line  to  the  left  extremity  of  the  horizontal  line  drawn. 
Use  the  divisions  of  the  cross-section  paper  to  read  tenths  of  7. 
Find  .43  x  ~.     What  change  must  you  make  to  read  4.3  x  7  ? 
or  43  x  7  ? 

3.  Make  the  vertical  line  7.4  in  length,  and  read  tenths  of 
7.4.     Find   approximately   .83  X  7.4.     Interpret   as   8.3  x  7. .4 
and  as  83  x  74. 

4.  On  the  oblique  line  of  problems  2  and  3  find  where  the 
vertical  distance  above  the  horizontal  line  is  5.     How  far  out 
is  this  on  the  horizontal  line  ?     This,  read  as  tenths  per  half- 

5  5 

inch,  represents  the  quotient  of  -  and  - —  respectively.     Find 

the  quotient  of  5.3  divided  by  7.     Find  5.3  divided  by  7.4. 

5.  Draw  a  semicircle  on  a  diameter  of  10  half-inches.     Xote 
that  the  perpendicular  at  any  point  on  this  diameter  is,  by 
plane  geometry,  a  mean  proportional  between  the  segments 
of  the  diameter.     Read  the  mean  proportional  between  1  and 
9,  as  the  vertical  line  drawn  at  the  point  on  the  diameter  4 
units  out  from,  the  center.    Read  the  mean  proportional  between 
2  and  8,  similarly  ;  between  3  and  7 ;  between  4  and  6. 

6.  Regard  the  preceding  circle  as  having  a  radius  10  quar- 
ter-inches.     Find   approximately,  from  it,   V19,  V36,  V51, 
V64,  V75,  V84,  V91,  V96,  V99,  and  V100.     These  are  the 
vertical  lengths  at  the  points  dividing  the  diameter  in  the 
ratio  19  to  1 ;  18  to  2 ;  17  to  3  ;  -  10  to  10. 


NUMBERS  OF  ALGEBRA  17 

7.  Take  a  circle  of  diameter  12  half-inches ;  from  it  ap- 
proximate Vll,  V20,  V27,  V32,  V35,  and_  V36.  Note 
that  V20=2V5,  V27  =  3V3,  and  V3^  =  4V2;  from  these 
approximate  V2,  V3,  and  V5.  See  diagram  on  page  72. 

8'.  On  the  preceding  circle  check  the  geometrical  fact  that 
either  side  of  a  right  triangle  is  a  mean  proportional  between 
the  whole  hypotenuse  and  the  adjacent  segment  of  the  hypote- 
nuse cut  off  by  the  perpendicular  from  the  vertex  of  the  right 
angle. 


CHAPTER   II 

APPLICATION    OF   ALGEBRA   TO   ARITHMETIC 

1.  Arithmetic  in  science.  —  In  science  definite  progress  is  usu- 
ally intimately  associated  with  the  arithmetical  investigation  of 
the  data  of  the  science.     Even  in  medicine,  biology,  sociology, 
and  chemistry,  arithmetic  plays  a  large  role  ;  in  physics,  astron- 
omy, and  engineering,  computation  is  absolutely  essential.    The 
student  of  mathematics  must  take  account  of  the  numerical 
side  of  mathematical  work,  constantly  applying  the  theoretical 
work  to  definite,  numerical  problems. 

The  accuracy  of  computed  results  in  scientific  work  depends 
upon  the  accuracy  of  the  observation  measurements  upon  which 
these  computations  are  based.  This  can  be  taken  ae  almost 
axiomatic.  Whenever  numbers  are  given  representing  meas- 
urements, the  computations  involving  these  numbers  should 
not  be  carried  beyond  the  point  which  these  measurements 
justify. 

2.  Arithmetical  application  of  algebraic  formulas.  —  Algebraic 
formulas  and  methods  can  frequently  be  applied  to  arithmeti- 
cal problems  with  a  great  saving  of  labor;  practice  with  nu- 
merical examples  is  absolutely  essential  for  success. 

The  student  is  expected  to  use  constantly  the  rules  given  in 
elementary  arithmetic  for  multiplying  and  dividing  by  5,  25, 
12.5,  334,  and  by  these  numbers  multiplied  by  integral  powers 
of  10.  Thus,  to  multiply  by  5,  multiply  by  5  as  ^°-,  annexing 
a  zero  and  dividing  by  2 ;  similarly,  25,  either  as  multiplier  or 
divisor,  is  considered  as  J-^,  and  so  with  the  other  aliquot  parts 
of  100  which  have  been  mentioned.  It  is  better,  in  general,  to 
use  these  rules  with  the  few  numbers  mentioned,  and  their 

18 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     19 

decimal  multiples,  rather  than  to  extend  this  work  to  other 
aliquot  parts  whose  occurrence  is  less  frequent. 

9350 

ILLUSTRATION.  —  25  x  374  is  written  4)37400 ;  2.5  x  375  is  written 
937.5 

4)3750.0  ;  173  H-  250  is  written  .173  x  4  =  .692  ;  75  x  375  is  written  : 
18750     (50  times  375,  dividing  375  by  2  and  extending  2  places.) 
9375     (25  times  375,  or  \  of  50  times.) 
28125     (75  times  375.) 

The  four  formulas  of  elementary  algebra  which  enjoy  the 
widest  use  are  undoubtedly  : 

(x  +  a)2  =  x*  +  2  ax  +  a2. 
(x  —  a)2  =  a;2  —  2  ax  +  a2. 
(x  +  a)(x  —  a)  =  a2  —  a2, 
(a;  +  a)(x  +  6)  =  x2  +  (a  +  b)x  +  ab. 

(x  +  a)(x  +  6)  gives  a  simple  rule  for  the  product  of  two 
"-teens,"  e.g.  19  x  17. 

Thus,  (10  +  a)(10  +  &)  =  102 +  (a+6)10  +  aft,  or  =10(10+  a+b)  +  ab. 

Put  into  words,  this  formula  states  that  the  product  of  two 
numbers  between  10  and  20  is  equal  to  the  whole  of  one  plus 
the  units  of  the  other ;  this  sum  is  to  be  multiplied  by  10 ; 
to  this  product  is  to  be  added  the  product  of  the  units. 

RULE.  —  To  find  the  product  of  two  "  -teens,"  add  the  whole 
of  one  to  the  units  of  the  other  and  annex  a  zero;  to  this  number 
add  the  product  of  the  units. 

19 
17_ 
260 
63 
323 

If  x  is  taken  as  20,  30,  40,  50,  ...,  the  corresponding  rule  for 
the  product  of  two  two-place  numbers  having  the  same  tens' 
digit  is  to  add  to  the  one  of  two  numbers  the  units  of  the 
other ;  the  sum  is  to  be  multiplied  by  the  tens'  digit,  and  a 


20  UNIFIED  MATHEMATICS 

zero  annexed  to  the  product ;  to  this  number  add  the  product 
of  the  units. 

Thus,  (37  x  35)  =  30  x  42  +  35  =  1260  +  35 

=  1295. 

Such  products  are  most  easily  found,  evidently,  if  the  two 

units'  digits  sum  to  10. 

87  X  83  =  8  x  9  x  100  +  21 

=  7221. 
64  x  66  =  6  x  7  x  100  +  24 

=  4224. 

In  mental  work  with  numbers  work  from  left  to  right,  and 
not  from  right  to  left,  dealing  first  with  the  numbers  of  greater 
significance. 

(x  4-  a)2  and  (x  —  a)2  are  particularly  useful  in  the  computa- 
tion of  squares  of  numbers  of  three  places  beginning  with  1 

or9. 

(10.7)2  =  100  +  2  x  10  x  .7  +  .49 
=  114.49. 

(11.3)2  =  121  +  6.6  +  .09  =  127.69, 
or  =  100  +  26  +  1.69  =  127.69. 

(1.57)2  =  2.25  +  .21  +  .0049, 
where  .21  is  obtained  as  1.5  x  .14  by  the  rule  for  the  product  of  two 

"-teens." 

(.97)2  _  (i.oo  -  .03)2  _  i  _  .06  +  .0009 

= .9409. 

(8.70)2  =  (io  _  1.3)2  _  100  -  26  -f  1.69  =  75.69. 

Frequently  it  is  more  convenient  to  use  these  formulas  re- 
arranged as  follows  : 

(x  +  a)2  =  x(x  +  a  +  a)  +  a2, 
(x  —  a)2  =  x(x—  a  —  a)  +  a2. 

Thus,  (84)2  _  100(100  -  16  -  16)  +  16" 

=  100(84  -  16)  +  162 
=  6800  -|-  256  =  7056. 
(25.7)2  =  20(25.7  +  5.7)  +  (5.7)2 
=  628  +  32.49 
=  660.49. 

(25.7)2=  25(25.7  +  .7)  +  .49  =  25(26.4)  +  .49 
=  660.0  (since  25  =  •»•£*)+  .49. 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     21 

The  square  of  any  number  between  25  and  75  is  obtained 
from  (x  +  a)2,  as  follows : 

(50  ±  a)2  =  2500  ±  100  a  +  a?  =  100  x  (25  ±  a)  +  at. 
Thus,  (37)2  _  2500  -  1300  +  169 

=  1369. 

RULE. —  To  find  the  square  of  any  number  between  25  and 
75 ;  find  the  difference  between  the  given  number  and  50;  add,  if 
the  given  number  is  greater  than  50,  or  subtract,  if  the  given 
number  is  less  than  50,  this  difference  from  25  and  annex  to  this 
two  zeros.  Add  to  this  number  the  square  of  the  difference. 

Thus,  (65)2  =(25  +  15)  x  100  +  152 

=  4225. 

For  numbers  between  75  and  150  the  squares  may  be  ob- 
tained as  100(100  -  a  -  a)  +  a2  or  (100)  (100  +  a  +  a)  +  a2, 
noting  that  100  —  a  or  100  -f  a  is  your,  given  number  whose 
square  is  sought. 

Thus,  1122  =  12,400  +  144. 

13.72  =  174.00  +  3.72  =  174.00  +  13.20  +  .49  =  187.69. 

Frequently,  of  course,  only  three  or  four  significant  figures 
are  desired,  and  the  methods  mentioned  give  the  significant 
figures  first. 

(x  +  d)(x  —  a)  may  also  be  used  for  squares,  thus : 

x2  =  (z  +  a)  (x—  a)  +  a2. 
(87)2  _  (87  +  13) (87  _  13)  +  132. 

(2.33)2  =(2.33+  .17)(2.33-.17)+.172 

=  5.4  +  .0289. 

(41.7)2  =(41.7  +  8.3)  (41.7  -  8.8)  +  (8.8)» 
=  50  x  33.4  +  8.32 
=  1670  +  68.89 
=  1738.89. 
(41.72)2  _  1738.89  +  (.04)  (41. 7)  +  .0004 

=  1738.89  +  1.6684  =  1741  to  units,  or  1740.6  to  tenths. 


22  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  Multiply  mentally  19  x  18,  17  x  15,  18  x  14. 

2.  Use  the  rule  given  above  to  give  the  table  of  18's  from 
18  x  11  to  18  x  19. 

3.  Multiply  mentally  12  x  13,  36  x  34,  45  X  45,  82  x  88, 
91  x  99. 

4.  Multiply  mentally  27  x  25,  34  x  32,  54  x  58,  92  x  98. 

5.  What  is  the  product  of  44  x  36  or  (40  +  4)  X  (40  —  4), 
58  x  62,  44  x  37  or  (40  +  4)  x  (40  -  3)  ? 

6.  What  are  the  first  three  figures  of  (114)2,  (107)2,  (131)*, 
and  (118)2  ?     Note  (114)2  is  12,800  +  142,  and  the  first  three 
figures  129 ;  in  (116)'-  to  13,200  you  must  add  (16)2,  which  in- 
creases the  first  13,200  to  13,400. 

7.  From  the  preceding  answers  in  6  write  the  first  three 
figures  of  (1.14)2,  (.107)2,  (1.31)2,  (H80)2. 

8.  Write  the  squares  of  9.7,  88,  940,  8.7,  and  9.2. 

9.  Approximately  how  much  greater  is  (9.71)2  than  (9.7)2? 
(88.2)2   than  (88)2?    (941)2  than  (940)2?    (8.75)'  than  (8.7)2? 
(9.26)2  than  (9.2)2? 

Note  that  (88.2)2  differs  from  (88)2  first  by  .4  x  88,  or  by 
a  little  more  than  35  units  ;  the  .04  is  usually  negligible. 

10.  Square  43,  47,  52,  63,  and  62  by  using  the  difference 
between  these  numbers  and  50  according  to  the  rule. 

11.  Using  the  preceding  answers,  square  4.3,  .47,  .052,  630, 
and  6.2. 

NOTE.  —  Use  common  sense  rules  to  determine  the  position  of  the 
decimal  point. 

12.  Using  the  formulas  for    (50  ±  a)2,  (a  ±  a)2,  (100  ±  a)2, 
write   the   following   25  squares.     Time  yourself   on  writing 
simply  the  answers ;  the  exercise  should  be  completed   in  6 
minutes. 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     23 

62*  = 
672  = 
822  = 

492  = 

13.   Using  the  results  of  the  preceding  exercise,  compute  to 
four  significant  figures  the  following  squares,  timing  yourself. 


572  = 

172  = 

242  = 

332  = 

632  = 

.  422  = 

392  = 

522  = 

872  = 

632  = 

592  = 

7I2  = 

432  = 

10.82  = 

212  = 

662  = 

982  = 

162  = 

922  = 

552  = 

57.12  = 

17.32  = 

42\52  = 

24.5*  = 

33.1*  = 

62.4'  = 

6732  = 

63.22  = 

39.72  = 

52.9*  = 

87.4*  = 

63.?  = 

59.22  = 

71.82  = 

82.9Z  = 

43.5-  = 

10.822  = 

21.42  = 

66.7Z  = 

1.922  = 

98.62  = 

16.6*  = 

92.62  = 

55.3'2  = 

49.82  = 

14.  Employing  the  formula  for  (#  -(-  a)  (x  +  6)  write  the 
following  products ;  the  exercise  should  be  completed  in  6 
minutes. 

16  x  19          22  x  24  32  x  38  51  x  52          66  x  64 
15  x  14          23  x  26  .  43  x  42  33  x  31          88  x  82 
13  x  18         24  x  29  46  x  44  27  x  24         97  x  93 

17  x  12          28  x  28  54  x  59  24  x  22          57  x  53 
16x18         36x33  82x87  17x13          79x71 

3.  Extraction  of  roots.  —  In  extraction  of  square  root,  the 
method  of  successive  approximation  should  frequently  be 
employed. 

Thus,  V179.63  >  13  and  <  14. 

Vl69  +  10.63  =  13  +  a,  wherein  a  must  be  a  number  such  that 
2  a  x  13  equals  approximately  10.6.  A  glance  shows  that  .4  x  13  equals 
5.2,  which  doubled  gives  10.4.  Hence,  (13.4)2  =  169  +  10.4  +  .16. 

(13.4)2  =  179.56,  or  179.63  -  .07. 
(13.4  +  a)2  =  179.56  +  2  a  x  13.4  +  a». 

a  now  is  less  than  .01  ;  hence,  a2  is  less  than  .0001  ;  a  is  to  be  a  number 
of  hundred ths  or  thousandths,  evidently,  so  that  2  a  x  13.4  is  approxi- 
mately .07  ;  a  is  roughly  .003,  slightly  too  large. 

(13.403)2  =  179.56  +  .0804  +  .000009 
=  179.6413. 


24 


The  rule  is  commonly  given  to  take  x  ±  -  '-  as  first  approxi- 

2  x 

mation  of  Vo?  ±a,  wherein  a  is  small  as  compared  with  x. 
The  process  illustrated  follows  this  rule,  but  suggests  thinking 
multiplication  instead  of  division.  Thus  in  the  square  root  of 
300,  as  V172  +  11,  approximately  -^  is  to  be  added  to  17 ; 
however  it  is  easier  to  think  34  X  a  =  11,  whence  a  =  .3,  or 
not  quite  .33 ;  trying  .32  (since  the  a2  term  is  to  be  added) 
gives  17.32  of  which  the  square  is 

289  +  10.88  (or  .32  x  34)  +  (.32)2  =  299.9829. 

Similarly,  V3000  =  V3025-25  =  V552-25  =55 -a,  wherein 
a  must  be  a  number  such  that  2  x  55  x  a  will  give  approxi- 
mately 25  ;  a  is  evidently  .2  to  one  decimal  place  or  .23  to 
two ;  54.77  is  correct  to  four  significant  figures  as  given. 

4.  Approximate  roofs.  —  Another  method  of  approximating 
square  root  is  to  divide  the  given  number  by  the  first  approxi- 
mation, then  to  use  the  arithmetic  mean  of  the  two  numbers 
as  a  second  approximation.  Thus,  179.63-^-13=13.82;  taking 

-\  o    I    -i  o  09 

-  as  the  approximate  root  gives    13.41  as  a  second 

a 

approximation.  179.63  -j- 13.41  gives  13.3952  and  the  average 
13.4026  is  within  .0001  of  the  correct  value. 

Similarly  the  cube  root  may  be  obtained.  Thus  in  179.63, 
5  is  the  first  approximation.  52  =  25  ;  179.63  -H  25  =  7.2 
nearly.  Taking  the  average  of  5,  5,  and  7.2  gives  5.7  as  second 
approximation  ;  (5.7)2=  32.49 ;  179.63  -=-  32.49  =  5.529  ;  the 
average  of  5.7,  5.7,  and  5.529  gives  5.643,  which  is  correct 
within  .001. 

PROBLEMS 

1.  What  is  the  approximate  square  root  of  1.26  ?  128  ? 

2.  Is  the  square  root  of  1.35  nearer  to  1.16  or  to  1.17  ? 
NOTE.     (1.16)2  =  1.32  +  .0256  and  (1.17)2  =  1.34  +  .0289. 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     25 

3.  By  successive  approximations  find  Vl.26  to  four  decimal 
places  and  compare  with  ordinary  process   of   extraction   of 
root. 

HINT.  —  Use  1.12  as  first  approximation. 

4.  Find  the  cube  root  of  1.26  by  this  process,  using  1.08  as 
first  approximation. 

5.  Find  the  approximation  to  one  decimal  place  of  the  square 
roots  of  65,  63,  8.30,  8.76,  and  27.32. 

HINT.  — Regard  8.30  as  (3  —  x)2,  whence  x  must  be  roughly  .12. 

6.  Write  the  square  roots  of  the  following  numbers,  correct 
to  2  decimal  places.     Time  yourself. 

9.9  35  65  140  200 

16.8  34  68  150  300 

17.2  37.2  78  125  10.4 

25.8  39.4  85  108  20.8 

28.  48  90  112  30.6 

5.  Synthetic  division  and  remainder  theorem.  —  The  ordinary 
process  of  division,  and  particularly  the  abbreviated  process 
by  detached  coefficients,  can  frequently  be  applied  to  com- 
putation. 3x2  +  4x  +  15 

x  —  2)3  x3  —  2  x2  +  7  x  —  5 
3x3 -6x2 
+  4x2 
+  4a2  —    8x 


+  15x 
+  15X-30 
+  25 

Notice  that  the  3  of  3  x2  has  been  written  three  times  ;  the  4  of  the  4  x2 
has  been  written  three  times  ;  the  15  of  the  16  x  has  been  written  three 
times.  Each  of  these,  3,  4,  and  15,  represents  not  simply  the  coefficient  of 
the  highest  remaining  term,  but  also  the  coefficient  of  the  corresponding 
term  in  the  quotient.  Further,  note  that  —  2  has  been  multiplied  by  3  x2, 
by  4x,  and  by  15,  and  the  product  with  sign  changed  has  been  added  to 
the  corresponding  term  of  the  dividend. 

x-2 

+  2)3-2+    7   -    5 

+  6+    8+30 

3  +  4+16(+26 


26  UNIFIED  MATHEMATICS 

This  represents  the  division  of  3x3  —  2x2  +  7  x  —  5  by  x  —  2  ;  the  —  2 
has  been  replaced  by  +2,  subtraction  has  been  replaced  by  change  of 
sign  and  addition  ;  the  remainders  have  not  been  written  but  are  merged 
in  the  quotient,  which  is  written  below. 

Divide 


-2)5+    0+    0-(-    8    +    7 

_  10  +  20  -  40    +64 

5 -10 +  20 -32  (+71 

Interpreted  :   ox4  +  8x  +  7  divided  by  x  +  2  gives  as  quotient 
6x3  —  10  x2  +  20  x  —  32  and  71  as  remainder. 

The  above  operations  may  also  be  interpreted : 

1.  3x3-2x2  +  7x-5=  (z-2)(3x2  +  4x+15)  +25. 

2.  5x«  +  8x  +  7  =  (x  +  2)(5x3-10x2  +  20x-32)  +  71. 

Substituting  in  (1)  x  =  2,  the  right-hand  member  reduces  to  the  con- 
stant term,  and  the  left-hand  member  is  the  original  expression  with  2 
substituted  forx.  Similarly  in  (2),  substituting  —  2  for  x  shows  that  the 
remainder  obtained  by  dividing  by  x  +  2  is  equal  to  the  original  expres- 
sion with  —  2  substituted  for  x. 

If  a  rational,  integral,  algebraic  expression  in  x  is  divided 
by  x  —  a  (a  may  be  negative),  the  division  being  continued 
until  the  remainder  does  not  contain  x,  the  remainder  is  equal 
to  the  original  expression  with  a  put  for  x.  Let  us  represent 
by  f(x)  an  expression  of  this  kind,  i.e.  an  expression  consisting 
only  of  the  algebraic  sum  of  positive  integral  powers  of  x 
with  constants  as  coefficients. 

Then,  you  can  divide  /(#)  by  x  —  a,  obtaining  a  quotient, 
Q(x~),  also  a  function  of  x,  and  a  remainder  which  does  not 
contain  x ;  by  definition  of  division,  f(x)  =  (x  —  a)  Q(x)  +  R ; 
substituting  a  for  a,  /(a)  =  (a  —  a)  Q(a)  +  It. 

/(a)  =  0  +  R, 
or  R  =/(a).     Q.  E.  D. 

The  common  "  check  by  nines  "  may  readily  be  proved  by 
the  remainder  theorem : 

When  12,738,  which  may  be  written 

1  x  10<  +  2  x  103  +  7  x  102  +  3  x  10  +  8, 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     27 

in  powers  of  10,  is  divided  by  10  —  1,  the  remainder  is  equal 
to  the  original  expression  with  1  put  for  10.  Hence  the  re- 
mainder when  12,738  is  divided  by  9  is  1  +  2  +  7  +  3  +•  8,  or 
21,  or  3  (since  21  divided  by  9  gives  3  as  remainder,  or  2  +•  1). 

Thus,  a  x  10"  +  b  x  10n-i  +  c  x  10n~2  +  •••  g  x  10  +  h  divided  by 
10  —  1  gives  a  -f  b  +  c  +  •••  +  g  +  h  as  remainder. 

Division  by  11,  10  +•  1,  gives  as  remainder  the  sum  of  the 
odd  coefficients  less  the  sum  of  the  even  coefficients,  counting 
from  units'  place  ;  a  sum  to  11  can,  of  course,  be  dropped  as  it 
occurs,  or  11  can  be  added  to  make  the  remainder  a  positive 
number. 

Illustrative  examples.  —  Dividing  by  the  method  of  detached 
coefficients  : 


1.   Divide  5x3  +  3a52  +  2a;  +  6bya;  —  2  and  by  x  +  2. 


2)5  +    3  +    2  +    6 

4-10  +  26  +56        5 
5  +  13  +  28(+62  =  (x-2)(5x2  +  13  x+ 28) +  62. 

x  +  2 

-2)5  +    3+    2  +    6 

—  10  +  14   -  32         5x3  +  3x2  +  2x  +  6 
6-    7  +  16(-26~  =  (x  +  2)(5x2-7x  +  16)  -26. 

2.   Divide  5326  by  98  and  102  ;  i.e.  by  100  -  2  and  100  +  2. 

5326  =  98  x  54  +  34. 

Note  that  the  500  is  subtracted  mentally,  and  since 
5x2  too  much  has  been  taken  away,  5x2,  or  10,  is 
added  to  the  remainder  32  ;  again,  4x2  too  much  has 

been  taken  away,  hence,  8  is  added  to  the  remainder. 
o4 

100  +  2 52 

5326  =  102  x  52  +  22. 

Here,   when  500  is  taken  away,  there  still  remains 

5  x  2  to  be  subtracted,  since  the  divisor  is  102. 
—  4 

~22 


28  UNIFIED  MATHEMATICS 

3.   Divide  127,384  by  96,  by  103,  by  124. 

96   1326        127,384  =  96  x  1326  +  88. 
4)127384 
4 

313 
12 
258 
8 

664 
24 
88  rem. 

i  A  little  practice  will  enable  one  to  perform  the  additions 

mentally,  writing: 

06      1326 
4)127384 
31 
25 
66 
88  rein. 

103       1236  103       1236 

-3)127384  or,  abbreviated,  -3)127384 

-3  24 

243  37 

-6  69 

378  76  rem. 

-9 

694 
-  18 

76  rem. 
120  +  4      1027 

—  4)127384  In  this  division  the  remainders  when  dividing 

—  4  by  120  must  be  noted  mentally  ;  thus,  330  less  240 

338  gives  90,  whence  98  as  remainder,  from  which  8  is 

—  8  to  be  subtracted.     Similarly,  900  less  840  gives  60, 
904               to  which  the  4  is  added   mentally,  and  28  sub- 

—  28  tracted  leaves  36  as  remainder. 
36  rem. 

A  number  of  two  places  ending  in  9  or  8  can  be  used  as  divi- 
sor in  short  division  form.     Thus, 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     29 

Think  60  in  127,  2,  with  remainder  7  ;  add 
2159  with  3  rem.       2  to   7,  giving  9  ;  93   by  60  gives  1,  with  re- 


59)127384  mainder  33  ;  33  +  1  =  34  ;  348  divided  by  60 
gives  5,  with  remainder  48  ;  48  +  5  =  53  ;  534 

by  60  is  9  less  6;  add  9  to  —  6  gives  +  3. 

88       1447,  Think  of  90  —  2  as  divisor. 

2)127384  127  -*-  90  =  1,  remainder  37  ;  37  +  2  =  39. 
J*_ 

393  393  -T-  90  =  4,  remainder  33  ;  33  +  8  (since  4  x  2  too 

8  much  has  been  subtracted)  =41. 

418  418  -*•  90  =  4,  remainder  58  ;  58  +  8  =  66. 
_8_ 

664  664  -f-  90  =  7,    remainder    34  ;    34  +  14  =  final     re- 

14  mainder,  48. 
48  renio 

PROBLEMS 

1.  Divide  2x3+3x2  —  9  a  —  Sbycc  —  1,  x—2,  x—  3,  x  +  l, 
x  +  2,  and  by  x  —  3,  employing  synthetic  division,  and  inter- 
pret results. 

+  2)2  +  3-    9--    5  _2)2  +  3-9-5 

+  4  +  14  +10  -4  +  2  +14 

2  +  7+    5(+    5  2-l-7(+  9     ' 

2af»  +  Sec*  -  9z  -  5  =  (x  -  2)(2a?  +  7a;+  5)  +  5. 
When  x  =  2,  2x*  +  3x*  - $x  -  5  =  +  5. 
2^  +  3^  —  9 a;  —  5  =(x  +  2)(2z2  —  x  -  7)  +  9. 
When  z=-2,2z3  +  3a;2-9a;-5  =  +  5. 

2.  Divide   x2  —  8xz  +  2 x  —  5   by  x  —  1,  x  —  2,  and  x  —  3. 
What  is  the  value  of  this  function  of  x  for  x  =  1,  x  =  2,  a;  =  3  ? 

3.  Divide  #4  —  3x2  —  18  by  x  —  2  and  by  x  —  4. 

4.  Divide  y?  -  1,289,000  by  x  -  100  and  x  -  110.     What  is 
the  value  of  x3  -  1,289,000  for  x  =  100  and  x  =  110  ? 

5.  Divide  af'  +  Sz2  —  1  x  —  21oy  x  —  Iby  synthetic  division. 

6.  Divide  2 or5  -  5x*  +  7*  -  10  by  x  -  2. 

7.  Divide  38,942  by  96  (as  100  -  4). 


30  UNIFIED  MATHEMATICS 

8.  Divide  38,942  by  59  (as  60  -  1)  by  short  division. 

9.  Divide  38J942  by  104  (as  100  +  4). 

10.  Divide  3  i*  -  2 t3  -  18  P  +  6  t  -  11  by  t  -  1.    Find  the 
quotient  and  remainder. 

11.  Find  the  value  of  z3— 179.63  when  x  =  5.6. 

12.  Find  the  square  root  of  321.62  by  repeated  division  and 
approximation,  as  indicated. 

13.  What  is  the  approximate  value  of  the  square  root  of 
(a)  145;  (6)  147;  (c)  150;  (d)  26;  (e)  2615? 

14.  Approximate   the   cube   root  of   126 ;    8.1 ;  27.4 ;  64.2 ; 
127  ;  218 ;  350 ;  520 ;  735 ;  and  1004. 

15.  Perform  the  following  divisions  by  synthetic  division ; 
give  quotient  and  remainder ;  time  yourself ;  20  minutes  is  an 
ample  allowance. 

a.  a?  -  3  x2  +  7  x  -  5  by  x  -  2. 

b.  2  a3  +  4  z2  —  7  a;  +  8  by  a  -  3. 

c.  3  x3  -  2  a;2  -  8  x  + 10  by  x  +  2. 

d.  4  ^-2^ +  8  a; -5  by  a  — .5. 

e.  &  —  7  a;2  +  8  x  —  5  by  x  —  2. 

f.  tf  -  18,700  by  x  -  25. 

g.  2  a?  -  100  xz  -  18,700  by  x  -  300. 
h.  y?  -  .2  x  -  .05  by  x  -  .8. 

t.   a5  -  3  ;r2  -  100  by  x-3. 
j.   Xs  -  2,000,000  by  x  -  140. 

16.  Divide  by  the  method  corresponding  to  synthetic  divi- 
sion ;  time   yourself ;   carry    the   division   to   four   significant 
risrures 

a.  1375  by  100  -  2  (or  98). 

b.  1375  by  60 -2. 

c.  4356  by  100 +  2. 

d.  8.736  by  5.9. 

e.  6248  by  7.9. 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     31 

6.  Percentage  of  error.  —  "When  any  measurement  of  physi- 
cal objects  is  given,  the  measurement  has  a  certain  limit  of 
accuracy,  determined  in  part  by  the  instruments  and  methods 
of  measurement  and  in  part  by  the  very  nature  of  the  thing 
measured.  In  measuring  the  distance  to  the  sun  from  the 
earth,  at  some  fixed  time,  the  measurement  may  be  given  as 
93,000,000  miles  ±  1,000,000  miles,  or  93,000,000  miles,  within 
a  million  miles  ;  the  thickness  of  a  watch  spring  may  be  meas- 
ured as  .014  inch  with  a  possible  error  of  one  thousandth  of 
an  inch,  or  .014  ±  .001  inch.  However,  from  the  point  of 
view  of  the  physicist  and  mathematician,  the  distance  to  the 
sun  is  more  accurately  given  than  the  thickness  of  the  watch 
spring,  for  the  percentage  of  error  —  ratio  of  possible  error  to 
measured  value  —  in  the  case  of  the  sun's  distance  is  slightly 
more  than  1  %  of  the  distance,  while  in  the  other  case  it  is 
more  than  7  %  of  the  thickness  of  the  spring. 

Every  number  which  represents  a  measurement  involves 
this  type  of  error.  Obviously,  in  any  computations  with  such 
numbers,  results  are  significant  only  within  limits  determined 
by  the  percentage  of  error. 

"7.  Significant  figures.  —  The  significant  figures  in  any  num- 
ber representing  a  measurement  are  those  which  are  given  by 
the  measurement,  and  do  not  include  those  initial  or  terminal 
zeros  which  are  determined  by  the  unit  in  which  the  measure- 
ment is  made.  The  terminal  zeros  in  93,000,000  are  not  sig- 
nificant figures,  as  the  unit  of  measurement  here  is  evidently  a 
million  miles  ;  as  the  measurement  can  be  made  to  one  further 
place,  the  distance  may  be  written,  in  "  standard  form,"  9.3  x 
107  miles  or  9.30  x  107  miles  in  which  only  significant  digits 
appear  in  the  first  factor  combined  with  powers  of  10.  In  the 
thickness  .014  inch,  the  initial  zero  is  not  a  significant  figure,  as 
it  is  apparent  that  the  measurements  are  made  in  thousandths 
of  an  inch ;  in  "  standard  form,"  this  is  1.4  x  10~3  inches. 

8.  Measurement  computations.  Products.  —  If  the  length  of 
a  rectangle  is  measured  with  an  error  of  less  than  1  %  of  its 


32 


UNIFIED   MATHEMATICS 


true  value,  and  if  the  breadth  is  given  absolutely,  the  true 
area  will  be  given  with  the  same  percentage  of  error  as  the 
length.  But  if  the  breadth  is  also  only  approximately  meas- 
ured, the  possible  error  in  the  area  obtained  as  the  product 

will  be  greater  than  if  only  one 
factor  involves  a  possible  error. 
The  graph  represents  the  right- 
hand  end  of  a  rectangle  whose 
length  and  breadth  are  measured 
as  21.5  cm.  and  12.2  cm.  respec- 
tively, where  it  is  understood 
that  the  measurement  only  pre- 
tends to  give  these  dimensions 
to  within  one  millimeter,  one 
tenth  of  one  centimeter.  The 
meaning  of  these  figures  then  is 
that  the  length  lies  between 
21.4  cm.  and  21.6  cm.,  and  the 
breadth  between  12.1  cm.  and 
12.3  cm.  The  first  is  an  error  of 
less  than  ^  of  1  %  and  the  sec- 
ond of  less  than  1  %.  The  un- 
certainty in  area  due  to  the  pos- 
sible error  in  length  is  indicated 
by  the  areas  at  the  right  end 
with  dimensions  .1  cm.  by  12.1 
cm.  or  .1  cm.  by  12.3  cm. ;  the 
uncertainty  in  area  due  to  the 
breadth  measurement  is  of  di- 
mensions .1  cm.  by  21.4  cm.  or  .1  cm.  by  21.6  cm.  The  area 
uncertainty  is  then  at  most  .1  cm.  by  (21.6  +  12.2)  cm.  or  3.39 
sq.  cm.  of  area.  This  error  may  evidently  affect  the  third 
figure  in  our  computation  of  the  area  and  hence  in  the  product 
the  figures  beyond  the  third  place  are  not  significant,  and  give 
no  real  information  concerning  the  actual  area  in  question. 
Note  that  the  area  as  the  product  of  12.2  by  21.5  is  262.30 


Itf      17      18      19     20     21 
Measurement  of  an  area 

A  rectangle  measured  as  21.5 
cm.  by  12.2  cm. 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     33 

sq.  cm.,  but  the  inaccuracy  of  measurement  of  the  length 
means  that  there  is  an  uncertainty  of  area  at  the  right-hand 
end  amounting  to  ±  1.22  sq.  cm.  (cm2.),  and  similarly  at  the 
top  an  uncertainty  of  ±  2.16  cm2.  ;  the  total  uncertainty  of 
area  amounts  to  more  than  3  cm2.,  and  should  be  given  as 
±  3.38  cm2.,  or  ±  3.4  cm2.  Commonly,  of  course,  the  measure- 
ments 12.2  cm.  and  21.5  cm.  mean  that  the  area  has  been 
measured  to  one  half  of  the  last  unit  given  ;  thus  this  area 
actually  falls  between  rectangles  of  dimensions  12.25  by  21.55 
and  12.15  by  21.45  ;  even  in  this  case  the  area  uncertainty 
is  greater,  by  precisely  similar  reasoning,  than  ±  1.5  cm2., 
and  is  approximately  1.7  cm2.  To  give  262.30  as  the  area 
of  this  measured  rectangle  is  giving  nonsense  in  the  last  two 
places  ;  it  should  be  given  as  262  or  262.3  ±  1.7  cm2. 

Let  k  and  k'  represent  measured  quantities  given  with  pos- 
sible errors  of  i  %  and  e  °J0  respectively,  e  and  i  being  assumed 
as  smaller  than  unity  (in  common  practice)  ;  the  absolute 

/          i  \ 
values  of  these  measured  quantities  lie  between  k  1  1  H  --  j 

and  "t-      and  between  *'1+ 


The  true  product  lies,  then,  between  kk'(  1-f-^-^H  --  —  —  ) 
and  kk'(  1—  —  t-?  -j-—-^_  V  in  other  words,  the  true  product 

\      100     looooy 


may  vary   by   l    —   from   the  computed   product  ;  —  —    is 

100  10000 

disregarded,  if  i  and  e  are  less  than  1,  since  the  fraction  is  less 

than  1  o/o  of  1  %  of  kk1.   In  the  graph  the  product      *f     x  kk' 

lUuOO 

is  represented  by  one  of  the  small  corner  squares  with  dimen- 
sions .1  cm.  by  .1  cm. 


Illustrative  example.  —  The  product  of  987  by  163  wherein  each  num- 
ber is  correct  to  within  £  a  unit  need  be  computed  only  to  the  fourth 
significant  figure  as  the  percentage  error  may  be  as  great  as  ^  of  1  %  +  ^ 
of  1  %,  since  £  in  987  parts  is  approximately  ^^  or  ^  of  1  %,  and  J  in 


34 


UNIFIED   MATHEMATICS 


163  is  greater  than  y^  or  T3ff  of  1  %.  The  error  in  the  product  may  be 
as  great  as  (^  +  T%)  of  1  %  or  275  of  1  %  ;  but  27ff  of  1  %  of  any  number 
certainly  affects  the  fourth  place  and  probably  affects  the  third  place  in 
the  number.  Hence  there  is  no  point  whatever  in  carrying  this  compu- 
tation beyond  four  places, 


987 

163 

98700 

59220 

2961 

160881 

163 

987 
1467 

130 
11 

160800  am. 


987 

163 

987  begin  with  100  x  987. 

692  take  6  x  98,  carrying  however  the  4  from  6x7. 
29  take  3x9,  carrying  the  2  of  3  x  8. 
160800 

987  +  J  987  -  i 

163+  \  163  -  \ 

160881  +  |(987  +  163)  +  $    160,  881  -  $(087  +  163)  +  $ 
=  161456$  =  160306| 


The  product  of  987  x  163  is  160,881  ;  987|  x  163£  gives  161,456$  ; 
986|  x  162£  gives  160,306$  ;  the  actual  area,  if  these  represent  dimensions 
of  a  rectangle  measured  to  three  significant  figures  lies  between  160,306$ 
and  161,456$.  In  practice  we  take  the  product  987  x  163  to  four  signifi- 
cant figures,  which  gives  the  area  slightly  more  accurately  than  our 
measurements  justify. 

9.  Abbreviated  multiplication  of  decimals.  —  The  abbreviated 
process  of  multiplication  applies  particularly  well  to  decimal 
fractions,  but  the  method  can  be  extended  to  integers  quite  as 
well.  To  find  .9873  x  .1346  correct  to  four  decimal  places. 

.9873 


begin  with  the  highest  digit  of  the  multiplier ;  first  x  fourth 

decimal  place  gives  fifth  decimal  place. 

continue  with  8x7  (second  x  third  place),  carrying  the  2  from 

8x3. 

take  4x8  (third  x  second  place)  carrying  3  from  4  x  7,  or  28, 

which  is  more  than  2  units  in  the  fifth  place. 

6  x  9,  or  54,  +  5  carried  from  the  6x8. 

It  assists  in  the  process  to  cross  out  the  last  upper  digit  as  it  is 

used  ;  thus  here  3  would  be  crossed  out  first,  then  7,  then  8,  and 

finally  9. 


9873 

7898 

395 

_59 
18225 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC     35 

If  a  check  is  desired,    multiply  again,  reversing  the   order  of  the 
factors  ;  thus : 

.1846 

.9873 

16614  begin  with  9x6. 

1477  take  8x4,  adding  5  from  the  8  x  6  product. 

129  take  7x8,  adding  3  from  the  7x4  product. 

6  take  3x1,  adding  2  from  the  3x8  product. 

.18225  Read  this  as  .1823  to  four  places. 

Obviously,  if  these  were  integers,  you  could  proceed  in  the  same  way, 
writing  the  final  product  with  four  zeros,  as  18,230,000. 


A  similar  abbreviation  can  be  effected  in  division  by  drop- 
ping each  time  the  last  figure  of  the  divisor  used,  and  using 
the^  remaining  part  of  the  original  divisor  as  new  divisor. 

Thus,  to  divide  .18225  by  .9873  or  by  .1846  you  proceed  as 
follows : 

^1846  .  .9873 

.18225  .1846).  18225 

9873  16614 

8352  1611 

7898  1477 


454  134 

395  129 

59  5 

59_  5 

Here  9873  is  used  as  the  first  divisor  ;  then  987  is  used,  but  to  the  partial 
product,  8  x  987,  is  added  the  tens'  digit  of  8  x  3,  the  digit  just  crossed 
out  ;  then  98  is  taken  as  divisor  and  to  the  product  is  added  the  tens'  digit 
of  4  x  7  (28  is  taken  as  giving  a  tens'  digit  of  3)  ;  then  9  is  used  and  5 
carried  over  from  6x8. 


10.  Percentage  effect  of  errors  in  divisor.  —  If  a  divisor  is 
known  to  be  too  large  or  too  small  by  a  definite  percentage  of 
itself,  the  quotient  will  be  respectively  smaller  or  larger  than 
the  correct  quotient,  for  small  per  cents,  by  approximately  the 
same  per  cent. 


36  IXIFIED   MATHEMATICS 

By  division,  -      —  =  1  —  i  +  i*  —  i3  -f-  i4  —  ••  • 
1  4-  i 

- 


For  values  of  i  less  than  .05,  i2  is  less  than  .0025,  or  ^  of 
1  % ;  i»,  ?'4,  and  t»  are  less  than  .000125,  .00000625,  and 
.0000003125,  respectively.  Hence  an  error  of  from  1  %  to 
5  %  of  excess  in  the  divisor  means  an  error  of  deficiency  vary- 
ing also  from  1  %  to  5  %,  within  \  of  1  %,  or  from  .99  %  to 
4.75%,  or  from  .9901%  to  4.7625%  in  the  quotient.  For 
values  of  i  between  \  of  1  %  and  1  %,  an  error  of  deficiency 
in  the  divisor  means  the  same  error  of  excess  in  the  quotient, 
within  -j-J-j  of  this  error.  The  meaning  in  physical  measure- 
ments of  these  results  is  that  when  the  divisor  is  correct  only 
to  the  third  significant  figure,  with  a  possible  error  of  £  to  1 
unit  in  the  third  place,  the  quotient  will  be  correct  to  about 
the  same  degree  of  accuracy. 

For  three-place  numbers  the  divisor  may  vary  from  100  to 
999.  The  possible  error  of  1  unit,  ±  ^,  means  that  100  must 
be  replaced  by  (100  ±  .5)  or  100(1  ±  .005)  and  999  by  999  ±  .5, 
or  approximately  999(1  +  .0005)  ;  the  quotient  will  vary  from 
1  —  .005  to  1  —  .0005  times  the  obtained  quotient.  Hence  the 
quotient  obtained  is  valuable  at  most  to  the  fourth  place,  and 
frequently  not  beyond  the  third  place. 

Illustration.  —  Given  that  76,430  is  divided  by  180;  what  variation 
ic  /iQft  i  KC\  AOA  a.  *n  tne  (luoti'eut  would  a  change  of  1  in  the  divisor 
produce  ? 

Suppose  that  instead  of  180,  170  should  have  been  used.  What  is  the 
error  in  the  "quotient  ?  180  —  1  —  180(1  —  .006),  the  error  in  the  divisor 
is  more  than  .5  %  and  less  than  .6  °Jo  of  the  divisor  ;  the  error  in  the 
quotient  is  no  more  than  2.5  and  no  less  than  2.1,  since  1  %  of  424.6  is 
4.246  and  .6  °/o  and  .5  %  are  respectively  2.5  and  2.1  ;  the  quotient  may 
be  taken  as  427.1,  whereas  426.9  is  obtained  by  actual  division.  Even  an 
error  of  \  a  unit  in  the  divisor  180  affects  the  third  place  in  the  quotient. 
In  obtaining  .5  %  and  .6  %  of  424.6,  there  is  no  point  in  carrying  the 
work  beyond  two  places  ;  the  values  show  that  the  error  is  between  2.1 


APPLICATION   OF  ALGEBRA  TO  ARITHMETIC     37 

and  2.5,  and  further  places  add  nothing  to  the  accuracy.  The  fraction 
3J5,  or  in  the  cass  of  the  \  unit  error  of  s^5,  might  just  as  well  be  used 
as  per  cents.  This  gives  in  the  latter  case  ^-^  of  424.6,  or  +  1.2  as  cor- 
rection, giving  425.8  as  quotient ;  the  actual  quotient  is  425.71)4. 

Do  not  carry  divisions  and  multiplications  beyond  the  degree  of  accu- 
racy warranted  by  the  data. 

Illustrative  examples. — You  can  multiply  3.14159  by  140.8  and 
obtain  the  result  numerically  correct  to  six  decimal  places.  But  if  the 
140.8  represents  the  diameter  of  a  circle,  measured  correctly  to  the  tenth 
of  an  inch  (or  of  a  foot,  or  of  a  meter)  the  product  of  140.8  by  3.14159 
gives  a  valuable  result  only  to  the  first  decimal  place  ;  the  circumference 
cannot  be  computed  correctly  to  any  further  percentage  of  accuracy 
than  that  with  which  the  diameter  is  measured.  The  area  can  be  com- 
puted here  with  any  meaning  only  to  four  significant  figures  ;  in  fact  an 
error  of  ±  .05  inch  in  the  diameter  makes  a  possible  error  of  ±  10 
square  inches  in  the  area.  It  is  convenient  to  write  the  products  from 
left  to  right,  dropping  work  beyond  the  second  decimal  place. 

140.8  140.8 

31  3.14159   the  .00059  is  of  no  use  as  it 


422 . 4  422 . 4  does  not  figure  i  n  the  prod  uct. 
20.1                                      14.1 

442.5  circumference  5.6 

1 
442.2  circumference 

For  most  practical  purposes  3^  is  sufficiently  accurate,  as  in  finding 
this  area,  wr2 : 

70.4 
70.4 
4928 
28 
4956  only  4  places  to  be  retained. 

4956 

3.14159 


14868 

496 

15576  area.  198 

5 

15567  area. 
Area  as  found  to  correspond  to  data,  15570. 


38  UNIFIED   MATHEMATICS 

PROBLEMS 

1.  The  distance  of  the  earth  from  the  sun  varies  between 
91.4  x  106  miles  and   94.4  x  106   miles.      The   length   of   the 
earth's  orbit  lies  between  circles  having  these  lengths  as  radii. 
Between  what  values  does  this  orbit  lie  ?      What  is  the  ap- 
proximate orbital  speed  of  the  earth  in  miles  per  hour  ? 

2.  The    mean    distance    of    the   earth    from    the    sun    is 
92.9  x  106  miles.     Compute  the  circumference  and  the  mean 
speed  and  compare  by  percentages  with  the  preceding. 

3.  Compute  the  speed  of  a  point  on  the  earth  due  to  the 
rotation,  taking  that  at  latitude  45°  the  radius  of  the  circle  of 
latitude  is  3050  miles.     Compare  the  rotational  speed  with  the 
revolutional  speed. 

4.  What  effect  on  the  computed  velocity  would  it  have  to 
take  365.25  instead  of  365  ?     How  would   you  correct   your 
division  for  365.25  as  divisor  after  having  obtained  the  quo- 
tient, dividing  by  365?  what  change  would  using  365.26  in- 
stead of  365.25  effect  in  the  computed  velocity? 

5.  The  distance  of  the  moon  from  the  earth  varies  between 
221,000  and  260,000  miles,  mean  238,000 ;  discuss  the  length 
of  the  path  of  the  moon  and  the  velocity  of  the  moon  which 
has  a  periodic  time  of  27.32  days. 

6.  A   man   whose   salary   is    S  3000   pays    $  480   for   rent. 
What  per  cent  is  this  of  his  salary  ?     Suppose  that  he  earns 
$  275  in  addition  to  his  salary,  what  per  cent  is  the  rent  paid 
of  his  income  ?     Compute  only  to  tenths  of  one  per  cent. 

7.  If  a  man  with  an  income  $  3275  pays  $  1100  per  annum 
for  food,  $  630  for  clothes,  $  240  for  life  insurance,  S  200  for 
"higher  life,"  and  saves  the  balance,  compute  his  budget  by 
per  cents. 

8.  Given  that  a  pendulum  of  length  I  cm.  makes  one  beat, 
one  oscillation,  in  t  seconds,  connected  by  the  relation, 

T 

'980' 
find  the  length  I  to  two  decimal  places  when  t  =  1. 


APPLICATION  OF  ALGEBRA  TO  ARITHMETIC       39 

9.   What  effect  on  /  does  a  change  from  980  to  981  pro- 
duce ?     What  decimal  place  in  I  would  be  affected  ? 

10.  What   error   would   the   use   of  3j-  instead  of  3.14159 
introduce  ? 

11.  The  number  n  of  vibrations  of  a  pendulum  of  length 
99.39  cm.  is  86400,  when  g  =  980.96 ;   g  is  the  acceleration 
due   to  gravity,  and  the   formula  for   the  number  of  vibra- 
tions is  given  by  the  formula, 

86400    la 

n= \/,> 

7T          »* 

or  for  the  seconds'  pendulum,  when  I  =  99.39,  g  =  980.96,  it 
is  n  =  86,400.  Suppose  that  at  the  top  of  a  mountain  (g  di- 
minishes) this  pendulum  of  length  99.39  loses  86  beats  per 
day,  what  is  the  approximate  percentage  of  loss  in  n  ?  The 
percentage  of  loss  in  g  is  approximately  double  this  since 

£       V2  * 

Vl  —  i  =  1  -  -  -.  What  is  the  approximate  loss  in  g? 
Take  86  as  TV  of  1  %  of  86,400. 

12.  Given  g  —  980,  I  =  50,  compute   n   in   the   formula  of 
problem   11.     What   maximum   effect   on  n  would   a   change 
from  I  =  50  to  /  =  50.5  cm.  produce  ? 

13.  Compute  the  weight  of  a  table  top,  hardwood,  dimen- 
sions correct  to  .05  foot,  top  48.1  x  36.4  x  2.1  inches.     Weight 
of  wood  48  pounds  per  cubic  foot. 

14.  If  a  table  top  similar  to  the  above  weighs  97  pounds, 
compute  the  weight  per  cubic  foot  of  the  wood. 

15.  If  water  weighs  62.4  pounds  per  cubic  foot,  compute 
the  specific  gravity  of  each  of  the  preceding  woods. 

wt.  of  cubic  foot  of  wood 


8  = 


wt.  of  cubic  foot  of  water 


16.  The  path  of  the  earth  is  approximately  a  circle  of  radius 
92.9  x  106  miles,  of  which  the  center  is  approximately  11  mil- 
lion miles  from  the  sun.  Compute  this  circumference  and 
compare  with  the  results  in  problems  1  and  2. 


CHAPTER   III 

EXPONENTS   AND   LOGARITHMS 

1.  Exponent  laws.  —  For  convenience  the  product  of  a  by 
itself,  a  x  a,  is  represented  by  a2,  a  x  a  x  a  by  a3,  ...,  and  a  •  a 
•  a  •  a  to  m  factors  by  am.  In  this  notation  m  is  called  the  ex- 
ponent and  a  the  base.  The  following  laws  evidently  hold : 

I.     am  '  a"  =  am+n. 

II.    —  =  am~n,  when  m  >  n. 
a" 

III.  (am)n  =  am'n. 

IV.  (a  •  6)m  =  am  •  bm. 

In  the  definition  as  given,  m  represents  the  number  of 
factors  and  is  assumed  to  be  a  positive  integer.  However,  it 
is  found  possible  to  define  am  for  all  real  values  (fractional, 
negative,  zero,  irrational)  of  m  so  as  to  have  the  resulting 
numbers  combine  according  to  the  four  laws  given  above. 

Thus,  a°  •  am  =  a0*"*  =  am,  if  Law  I  is  to  continue  to  hold ; 
hence,  a°  must  be  defined  to  equal  1,  since  multiplying  a 
number,  am,  it  gives  that  number.  To  be  justified  in  using  a 
zero  exponent  with  this  meaning  the  other  exponent  laws  must 
be  shown  to  Hold  when  either  m  or  n  is  zero,  but  in  II  only 
n  could  be  zero  at  this  point. 

For  a  negative  integer,  —  n,  if  Law  I  is  to  hold,  a  must  be 
defined  as  such  a  number  that  a"  •  a~n  =  a~n+n  =  a°  =  1 ;  hence 

we  define  a~n  as  the  reciprocal  of  a",  a~"  =  —     All  the  laws 

a" 

I  to  IV  can  be  shown  to  hold  under  this  extension  of  the 
meaning  of  a". 

40 


EXPONENTS  AND  LOGARITHMS  41 

p 
Similarly,  a q ,  if  Law  III  is  to  hold,  must  represent  a  number 

•^ 
which  raised  to  the  gth  power  equals  ap ;  aq  is  thus  defined  as 

the  qth  root  of  the  pth  power  of  a.     Taking  this  definition  of 

p 
a9,  Laws  I  to  IV  can  be  shown  to  hold  with  this  extension  in 

possible  values  of  m  and  n ;  p  and  q  are  assumed  to  be  integers. 

p        ™p 
For  fractional  exponents  a  fifth  law  is  introduced,  a*  =  a  "*. 

For  irrational  values  of  n,  an  is  defined  by  a  limiting  process. 
Thus,  d^*  is  defined  as  the  limit  of  the  series  a1-4,  a1-41,  a1-414,  ..., 
wherein  the  successive  rational  exponents  define  the  square 
root  of  2. 

The  operations  of  elementary  algebra  with  radicals  are 
made  subject  to  the  exponent  laws. 

Thus,     2V3  =  2-3*  =  (22)*  .  3*  =(22 .  3)*  =  12*  =  Vl2, 
by  successive  application  of  V,  III,  and  IV. 

The  operation  of  raising  to  a  power  indicated  by  am,  with 
m  integral,  is  called  involution.  The  inverse  operation  of 
finding  x  when  xm  is  given  equal  to  a  is  called  evolution. 

2.  Logarithms.  —  A  logarithm  is  an  exponent. 

The  relation  x  =  am  may  be  written  m  =  loga  x. 
m  is  the  exponent  which  applied  to  a  gives  x; 
m  is  the  logarithm  of  x  to  the  base  a. 

3.  Fundamental  laws  of  logarithms, 
a.   Logarithm  of  a  product. 

If  x  =  am  and  y  =  a", 
x  •  y  =  am  •  a"  =  am+n. 
logu  (x  -  y)  =  m  +  n  =  logu  x  +  loga  y. 


42  UNIFIED  MATHEMATICS 

In  the  language  of  logarithms  and  translated  into  ordinary 
language  this  theorem  is  as  follows  : 

I.    loga  (*•!/)  =  loga  x  +  Iog0  y ; 

in  words,  the  logarithm  of  a  product  is  the  sum  of  the  logarithms 
of  the  factors. 

b.  Logarithm  of  a  quotient. 

-  =  —  —  a*-»    loga  x-  =  m  —  n  =  loga  x  —  loga  y. 
y      an  y 

II.   loga  -  =  loga  x  —  loga  y ; 

in  words,  the  logarithm  of  a  quotient  is  the  logarithm  of  the 
dividend  minus  the  logarithm  of  the  divisor. 

c.  Logarithm  of  a  power. 

If  x  =  am,  xn  —  («m)n=  amn ;  loga  x"  =  m  •  n  =  n  loga  x. 
III.    loga  xn  =  n  loga  x ; 

the  logarithm  of  a  power  of  a  number  is  the  index  of  the  power 
times  the  logarithm  of  the  number. 

Since  our  exponent  laws  hold  for  all  values  of  m  and  ?i,  these 
theorems  hold  for  all  values  of  m  and  n. 

x  and  y  are  assumed  to  be  positive  numbers  and  for  compu- 
tation purposes  10  is  commonly  taken  as  the  base. 

We  assume  that  as  the  logarithm  increases  the  number 
increases.  This  can  be  readily  proved  from  the  fact  that 
10m  •  10"  =  10m+n ;  no  matter  how  small  the  n  is,  as  a  .positive 
quantity,  10"  is  greater  than  1.  For  n  any  positive  fraction, 

— ,  10"  represents  the  gth  root  of  the  pth  power  of  10,  wherein 

p  and  q  are  integers.  Now  10*  will  be  an  integer  10,  or  greater 
than  10,  and  the  <?th  root  of  this  integer  will  be  a  number 
greater  than  1  for  it  will  be  a  number  which  raised  to  the 
qth  power  equals  10,  100,  1000,  or  some  greater  number.  It 
could  not  be  less  than  1,  for  every  positive  integral  power  of 
a  number  less  than  1  is  also  less  than  1. 


EXPONENTS  AND   LOGARITHMS 


.000000000001 


1012 


1000000000000 


43 


EXPONENTIAL 
FORM 

COLUMN  OF 
NUMBERS 

COLUMN  OF 
NUMBERS    : 

COLUMN  OF 
LOGARITHMS 

log  .0000001)00001  v     =  -  12 


10-5          _ 

.00001 

log  .00001 

=  -5 

10-4     = 

.0001 

log  .0001 

=  -4 

10-3          = 

.001 

log  .001 

=  -3 

lO-2           = 

.01 

log  .01 

=  -2 

10-1      = 

.1 

log.l 

=  -  1 

10" 

1 

logl 

=  0 

101 

10 

log  10 

=  1 

102              = 

100 

log  100 

=  2 

103              - 

1000 

log  1000 

=  3 

104         = 

10000 

log  10000 

=  4 

105 

100000 

log  100000 

=  5 

log  1000000000000        =     12 


These  positive  numbers,  middle  columns,  are  arranged 
vertically  in  order  of  magnitude ;  the  exponents  (left)  or 
logarithms  (right)  also  are  arranged  vertically  in  order,  in- 
creasing from  — 12  (or  from  —  cc  indicated  by  dots  above)  to 
—  5,  to  —  4,  to  —  3,  —  to  0,  to  1,  •••  to  12,  •••  to  oo.  As  the  num- 
ber increases  the  logarithm  increases.  Placing  any  number, 
not  an  integral  power  of  10,  in  its  proper  place  as  to  magnitude 
on  such  a  diagram,  the  logarithm  has  for  integral  part  the 
logarithm  of  the  preceding  number  in  the  table.  Thus,  75.64 
falls  between  10  arid  100  and  its  logarithm  will  be  I4  ;  .07564 
falls  between  .01  and  .1  and  its  logarithm  will  be  —  2+,  mean- 
ing —  2  plus  some  positive  fraction. 


44  UNIFIED  MATHEMATICS 

4.  Logarithms,  characteristic  and  mantissa.  —  Any  two-place 
number  lies  between  10  and  100 ;  the  logarithm  will  lie  be- 
tween 1  and  2.  Any  four-place  number  lies  between  1000 
and  10,000 ;  the  logarithm  will  lie  between  3  and  4,  since  as 
the  number  increases  the  logarithm  increases.  The  fraction 
.07  is  greater  than  .01  and  less  than  .1 ;  the  logarithm  is  then 
greater  than  —  2  and  less  than  —  1. 

The  logarithm  of  any  number  between  1  and  10  is  a  frac- 
tion, expressed  commonly  as  a  decimal  between  0  and  1. 

The  logarithm  of  any  given  number  which  is  expressed  in 
decimal  notation  can  be  expressed  as  an  integer,  positive  or 
negative,  called  the  characteristic,  plus  the  positive  decimal 
fraction,  the  mantissa,  which  is  the  logarithm  of  that  number 
between  1  and  10,  having  the  same  succession  of  digits  as  the 
given  number.  Initial  zeros  are  not  included  in  the  succession 
of  digits. 

log  10"  Jc  =  log  10"  +  log  k 
=  n  log  10  +  log  k 
=  n  +  log  fc. 

Thus,  log  3.16229  =  .50000,  since  3.16229  is  the  approximate 
square  root  of  10. 

log  31622.9  =  log  (10)4(3.16229)  =  log  104  +  log  3.16229 
=  4  +  log  3.16229 
=  4.50000. 

log  .000316229  =  log  (10)-4(3.16229)  =  log  (10)~4  +  log  3.16229 
=  _  4  4-  .50000 

=  4.50000,  only  the  4  is  negative 

=  6.50000  -  10,  since  -  4  equals  6-10 

=  16.50000  -  20 

=  26.50000  -  30, 


The  minus  sign  over  the  4  indicates  that  only  the  charac- 
teristic is  negative  ;  the  alternate  forms  for  writing  a  negative 
characteristic  are  frequently  found  convenient  to  use,  particu- 
larly in  extracting  roots. 


EXPONENTS  AND  LOGARITHMS  45 

BULK.  —  TJie  logarithm  of  any  number  greater  than  unity  has 
as  characteristic  a  positive'  integer  (0  included}  tchich  is  1  less 
than  the  number  of  digits  to  the  left  of  the  decimal  point. 

Tfie  logarithm  of  any  decimal  fraction  has  as  characteristic 
the  negative  of  a  positive  integer  (0  not  included')  which  is  1 
greater  than  the  number  of  zeros  between  the  decimal  point  and 
the  first  significant  digit  (i.e.  digit  other  than  0)  to  the  right  of 
the  decimal  point. 

5.  Computation  of  logarithms.  —  Logarithms  are  actually  cal- 
culated by  a  series  derived  from  formulas  obtained  in  higher 
mathematics.  However,  a  simple  although  laborious  method 
of  computing  logarithms  approximately  may  make  the  signifi- 
cance of  the  logarithm  somewhat  clearer. 

210  =  1024,  220  =  1,048,576. 

Evidently,  106  <  220  <  107,  since  1,048,576  is   greater   than 
1,000,000  and  less  than  10,000,000. 
Extracting  the  twentieth  root, 


<  2  < 

or  10-30  <  2  <  10-35. 

Hence,  2  is  greater  than  10  with  an  exponent  .30,  and  less 
than  2  with  an  exponent  .35. 

240  =  (1,048,576)2  <  1,100,000,000,000  and  greater  than 
1,000,000,000,000  ;  280  <  1.21  x  1024  (240  by  240)  and  greater 
than  1  x  1024.  Whence  2100  <  1.33  x  1030  and  greater  than 
1  X  1030,  whence  1030  <  2100  <1031,  whence 

10-30  <  2  <  10-31,  or  Iog102  =  .30+. 

Similarly,  320  is  3,486,783,861,  or  greater  than  109  and  less 
than  1010;  hence,  log  3  lies  between  .45  and  .50,  the  computa- 
tion of  320  is  easily  made  by  using  34  =  81  and  320  =(81)5,  mul- 
tiplying each  time  by  81  instead  of  by  3  ;  the  partial  product 
by  the  1  does  not  need  to  be  rewritten. 


46 


UNIFIED  MATHEMATICS 


6.  Tables  of  logarithms.  —  The   exponents  or  logarithms  to 
the  base  10  of  all  numbers  up  to  100,000  have  been  computed 
by   methods   of    higher   mathematics ;    these   logarithms   are 
arranged  in  tabular  form  in  the  natural  order  of   the  corre- 
sponding numbers,  so  as  to   be   convenient  for   computation 
purposes.     Our  tables  give  the  mantissas  of  the  logarithms  of 
all  numbers  between  100  and  999 ;  by   the   insertion  of  the 
proper  characteristic  the  logarithms  of  all   numbers   having 
one,  two,  or  three  significant  figures,  i.e.  disregarding  initial 
and  terminal  zeros,  are  given  by  our  tables.     The  extension  of 
the  use  of  such  a  table  to  give  logarithms  of  numbers  having 
four  significant  figures  is  explained  in  the  next  section.     The 
method  applies  to  increase  in  a  similar  manner  the  scope  of 
any  table  of  logarithms  so  as  to  give  the  logarithms  of  num- 
bers having  at  least  one  more  significant  figure  beyond  those 
of  the  numbers  in  the  tables. 

7.  Logarithms.    Interpolation. — Tables  of  logarithms  give,  in 
general,  only  the  mantissas ;  the  characteristics  are  to  be  sup- 
plied according  to  the  rule  given  above. 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

This  section  of  a  logarithm  table  gives  the  logarithms  of  all 
numbers  with  three  significant  figures,  having  the  succession 
of  digits  beginning  20  or  21.  Note  that  the  differences  begin 
at  22  in  the  first  line  and  drop  to  19  in  the  second. 

These  logarithms  lie  between  2.3075  and 
2.3096  ;  they  increase  steadily  ;  we  assume  that 
they  increase  by  uniform  amounts,  which  we  see 
in  the  table  is  roughly  true  for  numbers  from  200 
to  219.  The  uniform  increase  of  these  logarithms 
of  203.1  to  203.9  is  fa  of  the  difference  between 
the  logarithms  of  203  and  204,  i.e.  2.1  of  the 
units  in  the  last  place  of  our  logarithms.  It 


log  203  =  2.3075 
log  203.1  =  ] 
log  203.2  =  I 

to 
log  203.9  =  j 

log  204  =  2.3096 


EXPONENTS  AND  LOGARITHMS 


47 


would  not  be  correct  to  increase  the  number  of  places  in  our  logarithms 
as  our  process  is  only  an  approximation  not  correct  to  further  places  even 
as  our  logarithms  are  only  approximations  to  four  decimal  places.  The 
logarithm  of  two  is  to  five  places  .30103,  to  ten  places  .3010299957,  to 
twenty  places  .30102999566398119521. 

log  203.0  =  2.3075 

203.1  =  2.3077 

203.2  =  2.3079 

203.3  =  2.3081 

203.4  =  2.3083 

203.5  =  2.3086 

203.6  =  2.3088 

203.7  =  2.3090 

203.8  =  2.3092 

203.9  ==  2.3094 


204.0  =  2.3096 


log  .203  =1.3075 
.2031  =  1.3077 
.2032  =  1.3079 
.2033  =  1.3081 
.2034  =  1.3083 
.2035  =  1.3086 
.2036  =  1.3088 
.2037  =  1.3090 
.2038  =  1.3092 
.2039  =  1.3094 
.2040  =  1.3096 


A   four-place   table   of  numbers,   with   logarithms    to    five 
places,  gives  simply : 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

203 

30750 

30771 

30792 

30814 

30835 

30856 

30878 

30899 

30920 

30942 

204 

30963 

30984 

31006 

31027 

31048 

31069 

31091 

31112 

31133 

31154 

The  appropriate  characteristic  must  be  inserted  by  the  com- 
puter. A  careful  examination  will  show  that  our  interpolation 
gives  an  incorrect  four-place  result  for  203.4,  with  an  error  of 
half  a  unit.  This  type  of  error  is  inevitable,  using  four-place 
tables.  In  general,  such  errors  tend  to  equalize  each  other; 
where  absolute  accuracy  to  four  places  is  necessary,  five-  or 
even  six-place  tables  must  be  used.  Even  with  four-place  tables 
it  is  evident  that  with  a  difference,  called  the  tabular  difference, 
of  21  to  24  the  normal  difference  will  be  2  units  and  1  to  4 
extra  units  will  have  to  be  distributed  in  the  addition. 

In  the  illustration  above  it  may  be  noted  that  log  203  as 
2.30835  does  not  inform 'us,  strictly,  as  to  whether  log  203  to 
four  places  is  2.3083  or  2.3084 ;  the  latter  is  the  case  here 


48  UNIFIED  MATHEMATICS 

since  if  log  203  is  found  to  further  places  than  five  the  fifth 
decimal  place  is  actually  a  5.  Whenever  the  five-place  loga- 
rithm is  a  terminal  5  which  has  been  obtained  by  increasing 
a  4  to  5  the  four-place  logarithm  Avould  not  be  increased  by 
one  unit;  thus,  log 2007  to  5  places  is  3.30255,  to  6  places 
log  2007  is  3.302547,  and  hence  to  four  places  log  2007  is 
3.3025.  In  some  tables  of  logarithms  a  terminal  5  due  to  an 
increase,  as  in  log  2007  =  3.30255,  is  marked  with  a  super- 
imposed negative  sign  or  other  mark. 

In  physical  problems,  measurement  to  three  places  permits 
of  computation  to  three  places  only  ;  the  fourth  place  by  inter- 
polation assures  accuracy,  in  general,  of  the  third  place. 

The  reverse  process  is  to  find  the  number,  given  the  loga- 
rithm ;  thus,  to  find  the  number  whose  logarithm  is  2.3088,  we 

see  that 

Diff.  21 

the  table  gives  log  203  =  2.3075  1 

and  log  204  =  2.3096  2 

The  tabular  difference  is  21  3 

The  given  logarithm  is  2.3088  4 

The  difference  between  it  and  the  5 

smaller  logarithm  nearest  to  it  is  13  6 


9 


2.1 
4.2 
6.3 
8.4 
10.5 
12.6 
14.7 
16.8 
18.9 


The  table  of  tenths  of  the  tabular  difference,  which  is  fre- 
quently given  in  tables  of  logarithms,  shows  that  13  is  nearest 
to  .6  x  21.  The  number  is  203.6.  Had  the  given  logarithm 
been  2.3087,  we  would  find  as  the  number  again  203.6,  since 
the  actual  difference  is  12,  which  also  is  nearer  to  .6  of  21  than 
to  .5  of  21.  Note  that  in  the  table,  and  in  the  difference 
table,  the  logarithm  or  part  of  the  logarithm  lies  always  to  the 
right ;  the  number,  or  part  of  the  number,  lies  to  the  left  or 
above. 

8.  Historical  note.  — Fundamentally  the  notion  of  logarithms 
is  intimately  connected,  as  we  have  shown,  with  the  notion  of 


EXPONENTS  AND  LOGARITHMS 


49 


exponents.     The  one-to-one  correspondence  between  exponents 
and  a  series  of  successive  powers  of  a  given  number  was  noted 


Exponents 
Numbers 


-4 

-3 

-2 

-  1 

0 

1 

•2 

8 

8 

4 

6 

6 

7 

8 

9 

10 

A 

i 

i 

* 

1 

2 

4 

16 

32 

64 

128 

256 

512 

1024 

Figure  from  Stifel's  Arithmetics  Integra,  1544 


many  years  before  logarithms  were  developed,  in  many  arith- 
metical textbooks  of  the  fifteenth  and  sixteenth  centuries. 
A  German  mathematician,  Stifel  (1486-1567),  published  a 
work  entitled  Arithmetica  Integra,  1544,  in  which  these  "  ex- 
ponentes,"  as  he  termed  them,  were  extended  to  the  left. 
Stifel  spoke  of  the  great  possible  use  of  such  series  for  compu- 
tation in  which  addition  would  replace  multiplication,  and  sub- 
traction replace  division  ;  but  he  developed  the  idea  no  further. 

In  1614  John  Napier,  Baron  of  Merchiston,  a  Scotchman, 
revolutionized  computation  processes  by  the  composition  of 
logarithmic  tables,  based  on  the  idea  of  the  comparison  of  two 
series  essentially  of  the  kind  indicated  above.  The  adoption 
of  10  as  the  base  of  a  logarithmic  series  was  due  to  a  friend  of 
Napier,  Henry  Briggs,  who  published  in  1617  the  decimal 
logarithms  of  the  first  thousand  numbers. 

In  recent  years  the  widespread  adoption  of  computing 
machines  which  carry  multiplications  and  divisions  to  fifteen 
and  twenty  places  is  somewhat  replacing  logarithms  in  the 
offices  of  great  insurance  companies  and,  to  some  extent,  in 
observatories. 

Logarithms.  ILLUSTRATIVE  PROBLEMS.  —  I.  Find  by  four- 
place  logarithms  : 


a.   203x137;          6. 

a.      log  203  =  2.3075 
log  137  =  2.1367 
log  product  =  4.4442 
product  =  278.1 


137' 


2003' 


d.  (203)3. 


By  interpolation  since  2,  the  difference 
found,  is  in  tenths  nearest  to  .1  of  16,  the  tab- 
ular difference. 


50  UNIFIED   MATHEMATICS 

6.     log  2.03  =  0.3075  log  2.03  =  10.3075  -  10 

log  137  =  2.1367  log  137   =    2.1367 

log  quotient  =  2.1708  1°S  quotient  =    8.1708  —  10 

quotient  =  .01482 

The  2  is  obtained  by  interpolation  since 
5,  the  difference  found,  is  nearest  to  .2  of 
29,  the  tabular  difference. 

c.    log  137  =  2.1367  d.  log  203  =  2.3075 
log  2.03=    .3075  3 

log  quotient  =  1.8292  log  (203)3  =  6.9225 
Quotient  is  67.58,  since  the  differ-  (203)3  =  8,363,000 

ence  5  is  in  tenths  nearest  to  .8  of  6, 
the  tabular  difference. 

If  these  numbers  represent  physical  constants,  obtained  by 
measurement,  our  computations  would  be  slightly  more  ac- 
curate than  the  measurements  warrant.  Even  in  prices,  in 
general,  these  results  would  be  sufficiently  accurate  for  pur- 
poses of  business.  Thus  137  pounds  of  cheese  at  20.3  ^  per  pound 
would  be  found  to  be  worth  $  27.81 ;  137  tons  of  mine-run 
coal  at  $  2.03  per  ton  is  worth  $  278.10,  with  a  possible  error 
of  10  ^ ;  1370  tons  of  coal  at  $  2.03  would  be  $  2781  with  a 
possible  error  of  50^,  which  is  negligible  in  business  of  this 
magnitude. 

II.   Find  by  logarithms,  four-place,  to  four  places : 

a.  203.8  x  137.5 ;     &.  ^^|;     c.  (203.8)3;     d.  (.02038)*. 


a.  log  203.8  =  2.3092  6.  log  20. 38  =  1.3092 

+  log  137.5  =  2.1378  -  log  13750  =  4.1378 

log  product  =  4.4470  log  quotient  =  3. 1714 

product  =  277,900  or  7.1714  _  10 

c.   log  203.8  =  2.3092  quotient  =  .001484 

3  By  interpolation  the  given  differ- 

log  (20.38)3  -  6.9276  ence  of  11(.1714  -  .1703)  is  nearest 

(20.38)3  =  8,464,000  in  tenths  to  .4  of  the  tabular  dif- 
ference, 29(.1732-  .1703). 


EXPONENTS  AND  LOGARITHMS  51 

d.  log  (.02038)5  _  |  i0g  .02038 

log  .02038  =  2.3092  8.3092  -  10 

3         °r        3 

5     6.9276  24.9276  -  30 

The  division  by  5  causes  trouble  because  of  the  negative  characteristic. 
To  avoid  the  difficulty,  write  6.9276  as  44.9276  -  50.  Dividing  this  by  5 
gives  8.9855  -  10  or  2.9855. 

(.02038)5  =  .09672. 

EXERCISES 

1.  By  chaining,  the  sides  of  a  rectangular  field  are  found  to 
measure  413.2  feet  and  618.4  feet.     Find  the  area  in  square  feet 
and  in  acres.     What  effect  upon  the  computed  area  does  an 
error  of  .1  of  a  foot  in  measurement   make  ?     Consider  this 
fact  in  making  the  computation,  not  assuming  that  these  lengths 
are  more  accurately  given  than  to  .1  of  one  foot. 

2.  Use     Hero's      formula,      A  =  Vs  (s  —  d)(s  —  b)(s  —  c), 

wherein  a,  6,  c  are  the  sides  and  ^       £-2  =  s,    to   compute 

2i 

the  area  of  a  triangle  whose  sides  are  413.2  feet,  618.4  feet, 
and  753.2  feet.  Discuss  errors  as  before.  Do  all  the  prelimi- 
nary computation  of  s,  s  —  a,  s  —  6,  s  —  c  before  looking  up 
any  logarithms. 

3.  Using   a  watch  with  a   second  hand,  time  yourself   on 
looking  up  the  logarithms  of  the  following  20  numbers ;  as  a 
preliminary  exercise  look  up  the  logarithms  of  these  numbers 
without   using   the   fourth   significant  figures,   not   requiring 
interpolating. 

log  314.6=  log       14.32  = 

log  813.2  =  log  1876000  = 

log  5.462=  log      81930  = 

log  .003468=  log     .08764  = 

log  .3085=  log      3.250  = 

log  769.9=  log       .7263  = 


52  UNIFIED  MATHEMATICS 

log  67870  =  log    200.4  = 

log  368.60=  log    399.8  = 

log    53.85=  log    210.4  = 

log    19.26=  log  .03899  = 

Twenty  minutes  should  be  the  maximum  time  on  this  list ;  practice 
until  you  can  do  all  20  with  interpolations  within  15  minutes  or  even  10 
minutes  ;  all  20  characteristics  should  be  written  before  you  begin  to  use 
the  tables. 

4.  Find  by  four-place  tables  the  logarithms  of  the  following 
numbers,   interpolating:    326,  .08342,   10,050,   .008766,  5499, 
3.482  x  106,  37.04,  290.40,  .9647,  38.55,  .06948,  3001,  2.777  x 
10-«,  784.4,  6,934,000,  5.341,  70.98,  .1237,  8462,  3740. 

Time  yourself;  the  20  should  not  take  more  than  12  minutes ;  the 
problems  should  be  written  down  before  the  table  is  used.  The  time 
includes  only  looking  up  the  logarithm  and  writing  it  down  ;  it  should 
not  include  copying  the  problem.  A  piece  of  blank  paper  may  be  placed 
alongside  of  each  column  and  the  logarithms  written  upon  the  paper  ;  by 
folding  the  paper  lengthwise  the  two  columns  may  be  placed  upon  the 
same  sheet.  Practice  with  timing. 

5.  Find  the  numbers  corresponding  to  the  following  loga- 
rithms —  no  interpolation  is  necessary.    Take  the  time  of  look- 
ing up  the  numbers  and  writing  these  in  a  prepared  form  as 
answers  ;  the  time  should  not  exceed  8  minutes. 

log  =  3.8414  log  =  1.0492 

log    .     =0.9996  log  =5.9063 

log  =2.6866  log  =2.9557 

log  =1.7482  log  =3.1732 

log  =6.3010  log  =6.2672 

log  =8.0086-10  log  =5.7832 

log  =4.2856  log  =1.1761 

log  =1.8837  log  =9.5024-10 

log  =9.3201-10  log  =2.4786 

log  =2.7789  log  =3.1673 

6.  Find  the  numbers  corresponding  to  the  following  loga- 
rithms, interpolating  wherever  necessary ;    time   yourself,  in 
looking  up  numbers,  first  writing   the  given  logs  in  column 


EXPONENTS  AND  LOGARITHMS       53 

form,  and  not  counting  that  time.  The  20  should  not  take 
more  than  15  minutes  ;  10  minutes  is  slightly  better  than 
average  time. 

log         =  3.5861  log  =  8.5418  -  10 

log         =5.6427  log  =2.0923 

log         =  1.4436  log  =  2.9844 

log         =4.7320  log  =3.3080 

log         =6.9428  log  =1.2818 

log         =  2.4415  log  =  7.8419  -  10 

=  6.4893  -  10  log  =  0.4630 

=  5.8662  log  =  1.7848 

=  1.5729  log  =  0.9618 

=  2.9990  log  =  1.7276 

7.  Time  yourself  on  finding  the  20  numbers  correspond- 
ing to  the  following  logarithms  : 

2.1120  1.2480  1.9462  2.7455  4.3925 

1.6150  5.4151  0.5132  0.0420  6.4105 

8.9312-10  3.5674  3.3808  1.2222  2.9213 

4.8990  1.1174  7.8973-10  1.3660  2.0621 

8.  Perform  the  following  computations,  using  logarithms  : 
a.  54.04x376.2;      b.  f^4-;  c.  (54.04)2(3.762); 


d.   ^5404;         e.  V54.04  x  376.2. 

9.   Find  the  volume  in  gallons  of  a  cylindrical  can,  18  inches 
in  diameter  and  30  inches  high.     (1  gallon  =  231  cubic  inches.) 

10.  Show  that  the  capacity  of  a  cylindrical  can  in  gallons 
can  be  written  as  £  of  1  %  of  d*h  +  2%  of  the  £  of  1  %  of  d*h, 
or  as  1.02  x  0.00^  X  dzh,  given  d  and  h  in  inches. 

11.  Find  the  volume  in  cubic  feet  of  a  silo  16  feet  in  diame- 
ter and  32  feet  high.     Compute  for  15.9,  15.95,  16.05,  and  16.1 
feet  in  diameter.  If  the  measurements  are  correctly  given  within 
.1  of  one  foot,  how  accurately  can  the  volume  be  given,  with  a 
16-foot  diameter  ? 


54  UNIFIED  MATHEMATICS 

12.  Find   approximate  value   of  2M,  2~so,  2^,  and  2~^  by 
logarithms. 

13.  Extract  the   cube   roots   of   2,  3,  4,  5,  6,  7,  and  8  by 
logarithms.     Multiply  the  value  of  2?  by  the  value  of  3*,  and 
compare  with  your  value  of  6s. 

14.  Given  a  pendulum  of  length  /  centimeters  and  time  of 
oscillation  t  seconds,  you  have  the  following  formula  connect- 
ing Z,  and  t  : 


Given  (  =  1,  compute  I;  given  I  =  60,  compute  t;  given  that 
t  =  1  but  that  instead  of  980  you  have  981,  compute  I. 

15-30.  Solve  the  problems  at  the  end  of  Chapter  II  using 
logarithms. 

9.  Compound  interest.  —  When  interest  is  added  to  the  prin- 
cipal at  the  end  of  stated  intervals  forming  a  new  principal 
which  is  to  continue  to  draw  interest,  the  total  increase  in 
the  original  principal  which  accumulates  by  this  process  con- 
tinued for  two  or  more  intervals  is  called  the  compound  inter- 
est on  the  original  principal. 

Let  P  represent  the  principal,  i  the  interest  rate  per  inter- 
val, and  n  the  number  of  intervals,  and  >S  the  amount  at  the 
end  of  n  intervals.  Given  interest  compounded  at  rate  t  per 
annum  for  n  years. 

At  the  end  of  1  year  you  have  P  -f  iP  =  P(l  +  1). 

At  the  end  of  2  years  you  have 

p(i  +  o  +  /P(i  +  0  =  P(i  +  O2- 

At  the  end  of  3  years  you  have 

P(i  +  i)2  +  *p(i  -f  i)2  =  P(i  +  O3- 


At  the  end  of  n  years  you  have          3  =  P(l  +  i)n. 


EXPONENTS  AND  LOGARITHMS  55 

Or  you  may  say  that  since  the  interest  for  1  year  in- 
creases the  principal  P  to  (1  +  *)P,  then  in  1  further  year  this 
new  principal  P(l  +  i)  will  be  increased  in  the  same  ratio, 
giving  P(l  +i)  x  (1  +  *)  or  P(l-f  i)2,  and  for  each  further  year 
the  factor  (1  +  i)  is  introduced.  Hence  the  amount  at  the  end 
of  n  years  is  P(l  +  i)n. 

If  interest  is  compounded  at  the  end  of  every  three  months, 

i  ? 

or  every  six  months,  you  substitute  for  i,  -  or  -  ,  and  for  n, 

4  n  or  2  n,  since  the  number  of  intervals  of  three  months  in  n 
years  is  4  n  and  of  six  months  is  2  n. 

f  j\mn 

The  formula  S  =  P(  1  +  —  )      is  used   for  an  interest  rate 

V       mj 
given  as  j  per  annum,  but  compounded  m  times  per  annum, 

at  rate  —  for  each  interval. 
m 

Problems  in  compound  interest  lend  themselves  to  solution 
by  logarithms. 

Given,  S  =  P(I  +  t)». 

log  S  =  log  P  +  log  (1  +  0". 
log  8  =  log  P  +  n  log  (1  +  i). 
log  P  =  log  S  -  n  log  (1  +  i). 


log  (1  +  i) 


Note  that  it  is  better  not  to  use  these  as  formulas,  memoriz- 
ing them,  but  rather  to  go  back  to  the  fundamental  relation, 
S  =  P(l  -+-  i)n.  Note  also  that  the  formula  holds  for  other 
than  integral  values  of  n  ;  thus  at  6  %  per  annum  the  interest 
on  the  amount  P  for  six  months  or  eight  months  is  defined  as 
P(l  +  .06)i  -  P  or  P(l  +  .06)1-  P,  respectively.  Hence  for 
n-f4  years  the  amount  would  be 


56  UNIFIED   MATHEMATICS 

PROBLEMS 

1.  Find  the  amount  of  $  1000  at  interest  4  %   annually, 
compounded  for  20  years.    Find  the  amount  when  compounded 
semi-annually  at  a  nominal  rate  of  4  %  per  annum,  i.e.  2  % 
semi-annually. 

2.  In  how  many  years  will  money  double  itself  at  4  %, 
5  %,  6  °lo  interest,  compounded  annually  '.' 

3.  Given  that  at  the  end  of  20  years  $  1000  amounts  to 
$  1480,  what  is  its  approximate  rate  of  interest  ? 

4.  Given  that  at  5  %  interest,  compounded  annually,  $  1000 
amounts  to  $  1480,  what  is  the  approximate  number  of  years  ? 

5.  Find  the  compound  amount  of  $  1400  at  5  %  interest, 
compounded  semi-annually,  for  10  years,  11  years,  12  years, 
up  to  20  years. 

6.  If   $  100  is  left  to  accumulate  at   3  %   interest,  com- 
pounded annually,  what  will  it  amount  to  in  100  years  ?    Solve 
by  logarithms.     What  amount  put  at  3  %  interest  will  amount 
in  100  years  to  $  1,000,000  ?     What  is  the  present  equivalent 
of  $  1000  to  be  paid  at  the  end  of  100  years,  money  worth  3  °J0 , 
compounded  annually  ? 

7.  Solve  problem  6  for  4%,  5%,  and  6%  interest,  com- 
pounded annually.      For   6  °J0    per  year,   compounded    semi- 
annually,  for  50  years. 

8.  Benjamin  Franklin,  who  died  in  1790,  left  1000  pounds 
to '"the  town  of  Boston"  and  the  same  to  the  city  of  Phila- 
delphia.    His  will  directed  that  this  amount  should  be  loaned 
at  interest  to  young  artisans,  and  thus  accumulated  for  100 
years  until  the  principal  should  have   increased   to   130,000 
pounds.     He  directed  further  that  at  that  time  the  major  por- 
tion of  this  amount  should  be  expended  for  some  public  im- 
provement and  the  residue  left  to  accumulate,  similarly,  for 
another  hundred  years.     What  rate  of  interest  did  Franklin 
assume  that  his  money  would  earn  ?     In  Boston  the  amount, 


EXPONENTS  AND  LOGARITHMS       57 

$  5000  approximately,  accumulated  to  about  $  400,000.  Find 
the  average  rate  of  interest  earned  annually.  Assuming  that 
$  5000  was  kept  aside  in  1891,  as  directed,  what  will  this 
amount  to  in  1991,  compounded  at  4  %  annually  ? 

9.   Find  the  amount  at  the  end  of  200  years  of  $5000, 
interest  at  4  %,  5  $>,  and  6  %,  compounded  annually. 

10.  If  a  business  doubles  its  capital,  out  of  earnings,  in 
12  years,  what  rate  of  interest  on  capital  invested  does  this 
represent  per  year  ?     If  in  20  years  the  capital  is  doubled,  find 
the  rate  of  interest  earned. 

11.  The  United  States  has  increased  in  population  from  7.2 
million  in  1810  to  101.1  million  in  1910 ;  find  the  approximate 
rate  of  increase  per  year,  and  for  each  ten-year  period.     Com- 
pare with  the  figures  on  page  65. 

12.  The  city  of  New  York  increased  in  population   from 
120,000  to  4,769,000  in  the  interval  from  1810  to  1910.     Com- 
pute the  average  annual    rate  of  increase,  using  the  formula, 
120,000(1  +  i}m  =  4,769,000.      Compute  the  average  ten-year 
increase  and  compare  with  the  actual  statistics  on  page  66. 


CHAPTER   IV 

GRAPHICAL   REPRESENTATION   OF  FUNCTIONS 

1.  Functional  relationships.  Expressions  of  the  form  3  x  -\-  5, 
ax  +  b  are  called  linear  or  first  degree  functions  of  the  variable 
x ;  in  elementary  algebra  such  expressions  have  been  combined 
according  to  the  fundamental  operations  and  subject  to  the 
laws  given  in  a  preceding  chapter.  Further,  some  atten- 
tion is  given  in  elementary  algebra  to  expressions  of  the 
form  ax*  +  bx  -f  c,  the  general  quadratic  function  of  x,  and 
expressions  involving  higher  powers  of  x.  The  expression 
axn  +  bxn~l  +  •••  is  called  an  algebraic  function  of  x  of  the 
nth  degree  when  n  is  a  positive  integer  and  the  coefficients 
a,  b,  •••  are  constants.  This  represents  of  course  a  number  for 
any  value  of  x. 

F(x),  G(x),  <t>(%),  A(#),  •••  are  methods  of  representing 
functional  relationships  ;  F(x),  (read  F  of  x  or  F  function  of  x), 
means  that  this  expression  assumes  various  values  as  x  varies, 
these  values  being  determined  by  some  law.  In  the  equation, 
y  =  3  x  +  5,  y  is  explicitly  given  as  a  function  of  x ;  y  is  here 
a  linear  function  of  x.  In  the  equation,  y  =  xz  +  4  x  —  5,  y  is 
an  explicit  function  of  x ;  as  a;  varies,  so  does  y.  In  x2-^  yz=  25, 
as  x  ta.kes  on  different  values  so  does  y,  but  one  must  solve  for 
the  corresponding  values  of  y.  Here  y  is  called  an  implicit 
function  of  x. 

When  two  variable  quantities  are  so  related  that  the  varia- 
tion of  one  of  these  depends  upon  the  variation  of  the -other, 
either  is  said  to  be  a  function  of  the  other.  Thus  the  pro- 
duction of  wheat  in  the  United  States  from  1900  to  1915  is  a 
variable  quantity  depending  upon  the  year  of  production. 
The  height  of  a  given  tree  is  a  function  of  its  age ;  to  each 
number  expressing  in  any  convenient  unit  of  time  the  age  of 

58 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS  59 

the  tree  corresponds  a  given  number  expressing  the  height 
of  the  tree.  Similarly  the  weight  of  a  tree  is  a  function  of 
the  age  of  the  tree.  This  type  of  relationship  cannot  be  ex- 
pressed algebraically.  It  may  be  exhibited  by  the  two  series 
of  numbers,  or  it  may  be  expressed  graphically. 

2.  Graphical  representation  of  statistics.  —  Since  two  variable 
quantities  are  to  be  represented,  two  sets  of  numbers  must  be 
indicated ;  this  could  be  done  by  placing  the  two  sets  upon 
two  lines  straight  or  curved,  drawn  parallel  to  each  other. 
This  is  the  form  used  upon  grocers'  scales  wherein  the  vari- 
ables of  weight  and  corresponding  price  are  placed  upon  con- 
centric circular  arcs ;  corresponding  numbers  are  cut  by  the 
pointer. 


Series  of  corresponding  numbers  graphically  represented 

It  is  commonly  more  convenient  to  place  the  two  scales  for 
representing  the  two  variable  quantities  upon  two  lines  per- 
pendicular to  each  other.  Upon  the  following  figure  the  tem- 
perature and  barometric  pressure  are  indicated  by  the  diagram 
for  the  week,  March  4-11,  1918,  at  Ann  Arbor,  Michigan. 


60 


UNIFIED  MATHEMATICS' 


Temperature  and  barometric  chart  by  moving  pointer 
The  sharp  break  in  barometer  curve  corresponds  to  a  violent  rainstorm. 

The  horizontal  displacement  of  any  point  on  either  graph,  located  by 
the  vertical  rulings,  indicates  the  time  of  the  observation  ;  the  correspond- 
ing temperature  or  pressure  is  indicated  by  the  vertical  displacement. 

ILLUSTRATIVE  EXERCISES 

1.  Production  and  price  of  wheat  in  the  U.  S.  from  1895  to 
1916  are  given  in  statistical  form  and  graphically. 


TEAR 

PRODUCTION 

EXPOBTS 

PRICE 

YEAR 

PRODUCTION 

EXPORTS 

PRICK 

Millions  of  Bushels 

Cents 

Millions  of  Bushels 

Cents 

1895 

467 

126 

50.9 

1907 

634 

163 

87.4 

1896 

428 

145 

72.6 

1908 

665 

114 

92.8 

1897 

530 

217 

80.8 

1909 

737 

87 

98.6 

1898 

675 

222 

58.2 

1910 

635 

69 

88.3 

1899 

547 

186 

58.4 

1911 

621 

80 

87.4 

1900 

522 

216 

61.9 

1912 

730 

143 

76.0 

1901 

748 

235 

62.4 

1913 

763 

146 

79.9 

1902 

670 

203 

63.0 

1914 

891 

332 

98.6 

1903 

638 

121 

69.5 

1915 

1,026 

243 

91.9 

1904 

552 

44 

92.4 

1916 

640 

160.3 

1905 

693 

98 

74.8 

1917 

1906 

735 

148 

66.7 

Statistics  from  the  Yearbook  of  the  U.S.  Department  of  Agriculture 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     61 


Graphical  representation  of 
(broken  line) 


wheat  production  (continuous  line)  and  price 
in  the  United  States,  1895-1917 


Note  that  the  graphical  form  of  these  statistics  brings  out  several  points 
of  interest.  In  the  first  place  the  maximum  price  paid  for  wheat  in  the 
interval  is  immediately  found,  and  so  also  the  minimum  price  of  51  ^ 
(50.9^)  in  1895.  Further,  the  diagram  shows  very  pointedly  that  a 
large  production  under  normal  circumstances  is  accompanied  by  a  fall  in 
price,  and  an  immediate  diminution  of  production.  In  1917,  under  war 
conditions,  both  production  and  price  increased  greatly. 

2.  The  weight  of  water  per  cubic  foot,  or  60  pints,  is  62.4 
pounds.  For  cylindrical  vessels  filled  to  a  height  of  12  inches 
the  weight  for  an  area  144  square  inches  in  the  base  would  be 
62.4  pounds  ;  for  72  square  inches  in  the  base  the  weight  of 


62 


UNIFIED  MATHEMATICS 


water  would  be  31.2  pounds  ;  for  0  square  inches  the  weight  is 
0  pounds.  On  coordinate  paper  represent  square  inches  of 
base  on  the  horizontal  line,  taking  1  major  division  to  repre- 
sent 10  square  inches,  and  represent  weight  on  the  vertical 
line. 


Weiyltt 


0     10    20    30    40    50    60    70    80    90   100  110  120  130 140 
Square  Inches  in  base 

Weight  of  water  in  cylindrical  vessels  with  varying  base  when  filled  to  a 
height  of  12  inches  or  10  inches 

Note  that  the  weight  of  12  inches  of  water  in  a  vessel  with  a  base  con- 
taining 50  square  inches  is  21.5  pounds,  and  conversely  if  a  cylindrical 
vessel  contains  21.5  pounds  in  12  inches  of  height,  the  base  contains  50 
square  inches,  and  similarly,  of  course,  for  other  values. 


PROBLEMS 

1.  Plot  the  temperature,  as  vertical  lengths,  and  the  time, 
by  hours,  as  horizontal  lengths,  for  24  hours. 

2.  Plot  the  contents    in   pints    of   cylindrical    vessels    12 
inches  in  height,  with  varying  bases ;  take  that  with  base  144 
square  inches,  the  capacity  is  60  pints ;  with  72  square  inches 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS  63 

in  the  base,  30  pints  ;  with  0  base,  0  pints.  The  straight  line 
joining  these  points  can  be  used  to  give  the  base  in  square 
inches  of  any  cylindrical  vessel  whose  capacity  for  a  height  of 
12  inches  is  known.  What  would  be  the  base  of  a  vessel  that 
contains  10  quarts  when  filled  to  the  height  of  12  inches? 

3.   Plot  cubic  inches  against  pints,  taking  1728  cubic  inches 
as  60  pints. 

INCREASE  IN  VOLUME  WITH  TEMPERATURE  INCREASE 

As  liquids  are  heated  the  volume  changes,  generally  increasing  ;  thus 
water  increases  in  volume  when  heated  except  between  0  and  +  4°  C. 
Given  1000  cu.  cm.  of  water  at  4°  C.  and  1000  cu.  cm.  of  mercury  at  0°  C., 
the  volume  at  other  temperatures  is  given  by  the  following  table  : 


TEMPERATURE 

VOLUME  OF  WATER 

VOLUME  OF  MEKCTRY 

0° 

1000.13 

1000.00 

1° 

1000.06 

1000.2 

4° 

1000.00 

1000.9 

8° 

1000.13 

1001.4 

10° 

1000.27 

1001.8 

15° 

1000.87 

1002.7 

20° 

1001.77 

1003.6 

25° 

1002.94 

1004.5 

30° 

1004.35 

1005.4 

35° 

1005.98 

1006.3 

40° 

1007.82 

1007.2 

4.  Plot  the  increase  above  1000  cu.  cm.,  or  decrease,  in  cu. 
cm.  in  volume  of  the  water,  using  1  half-inch  for  1°  on  hori- 
zontal axis  and  1  half-inch  for  1  cu.  cm.  on  the  vertical  axis. 
Note  that  by  adding  1000  to  the  given  readings,  actual  volumes 
can  be  read. 

5.  Plot  the  same  curve  for  the  increase  in  volume  of  the 
mercury.     It  is  evident  that  the  increases  in  volume  of  the 
mercury  are  approximately  proportional  to  the   increases   in 
temperature. 


64 


UNIFIED  MATHEMATICS 


STATISTICS  ON   WEKJHT  AND  HKKMIT 

From  an  investigation  of 'the  statistics  giving  characteristics  of  a  group 
of  over  200,000  men  and  130,000  women,  the  following  facts  are  obtained 
on  average  height.  The  facts  are  given  for  groups  of  1000. 


HEIGHT 

FRK^I  EN<  v  mi  No.  IN 
GBOirp 

Men 

Women 

4'  9" 

0 

1 

4'   10" 

0 

4 

4'   11" 

0 

10 

5'  0" 

2 

40 

5'  1" 

2 

55 

5'  2" 

5 

107 

5'  3" 

12 

135 

5/  4" 

30 

184 

5'  5" 

55 

167 

5'  6" 

90 

134 

5'  7" 

127 

83 

6'  8" 

169 

48 

5'  9" 

145 

18 

5'  10" 

147 

8 

5'  11" 

104 

3 

6'  0" 

66 

1 

6'  1" 

22 

0 

6'  2" 

11 

0 

6'  3" 

3 

0 

6'  4" 

1 

0 

WEIGHT  (TO  XKARKST 

I.VIKi.KK    IN    5  OR  0)  OK 

MEN  ;  AGKS  3/>-#J  ; 
HEIGHT  5'  lu" 

FREyfENCY   OR 
Nl'MIIKR    IN 

Qmuvr 

125 

4 

130 

14 

135 

33 

140 

60 

145 

78 

150 

114 

155 

95 

160 

106 

165 

90 

170 

87 

175 

72 

180 

59 

185 

48 

190 

37 

195 

25 

200 

32 

205 

12 

210 

14 

215 

4 

220 

8 

225 

3 

230 

1 

In  any  such  group  the  number  of  individuals  having  any  given  char- 
acteristic is  called  the  frequency  corresponding  to  the  given  characteristic. 

6.  Plot  the  frequency  curve  of  heights  of  men  and  women, 
taking  ^  inch  as  corresponding  to  1  inch  of  height  on  the 
horizontal  axis  and  taking  ^  inch  vertical  for  30  individuals. 
This  curve  represents  very  nearly  what  is  termed  a  normal 
symmetrical  distribution. 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     65 


7.  Plot  the  frequency  curve  for  weights  of  men  between 
35  and  39. 

AGRICULTURAL  STATISTICS 

In  the  Yearbook  of  the  Department  of  Agriculture  statistics  of  produc- 
tion and  prices  of  standard  crops  and  farm  products  are  given,  covering  a 
period  frequently  of  50  years.  Use  this  Yearbook  to  obtain  the  data  for 
the  following  curves : 

8.  Plot  the  curve  showing  the  production  of  corn  in  the 
United  States  from  1866  to  the  present  time.     Use  200,000,000 
bushels  as  a  vertical  unit,  taking  ^  inch  as  the  unit ;  take  one 
year  as  y1^  of  an  inch. 

9.  Plot  prices  on  the  diagram  of  8,  using  a  right-hand  scale. 

10.  Plot  similarly  statistics  for  the  amount  and   price   of 
sugar  produced  in  the  United  States. 

11.  Plot  the  average  price  in  the  United  States  of  eggs  by 
months  for  the  current  year ;  plot  butter  prices  similarly. 

POPULATION  STATISTICS 

The  population  statistics  of  the  United  States  by  10-year  intervals  as 
given  by  the  1914  Statistical  Atlas  of  the  U.  S.  Bureau  of  Census  are  as 
follows  : 


DATE 

U.S. 
(Millions) 

NEW  YOKK 
(Thousands) 

TEXAS 
(Thousands) 

1790 

3.9 

340 

1800 

5.3 

589 

1810 

7.2 

959 

1820 

9.6 

1,373 

1830 

12.9 

1,919 

1840 

17.1 

2,429 

1850 

23.2 

3,097 

213 

1860 

31.4 

3,881 

604 

1870 

38.6 

4,383 

819 

1880 

50.2 

5,083 

1,592 

1890 

63.0 

6,003 

2,236 

1900 

77.3 

7,263 

3,049 

1910 

101.1 

9,114 

3,897 

66 


UNIFIED  MATHEMATICS 


12.  Plot  the  curve  of  population  of  the  United  States. 

13.  Plot  the  population  curve  for  Michigan,  and  estimate 
the  population  for  the  5-year  periods. 


DATE 

MICHIGAN 
(Thousands) 

DATE 

NEW  YOBK  CITY 
(Thousands) 

DATE 

MlCHH.  \  v 

(Thousands) 

XF.W  YOKK  CITY 
(Thousands) 

1837 

175 

1790 

49 

1864 

804 

1800 

79 

1870 

1,184 

1V478 

1810 

120 

1874 

1,334 

1820 

152 

1880 

1,637 

1,912 

1830 

242 

1884 

1,854 

1890 

2,094 

2,507 

1840 

212 

391 

1894 

2,242 

1900 

2,421 

3,437 

1845 

303 

1904 

2,530 

1910 

2,810 

4,769 

1850 

398 

696 

1854 

507 

1860 

749 

1,175 

14.  Plot  the  population  curve  of  New  York  City.     What 
has  been  the  average  rate  of  increase  for  10-year  intervals  and 
for  yearly  intervals,  approximately,  since  1810?     Note  that 
this  requires  the  solution  of  the  equations  120(1  +  /)10  =  4769, 
and  120(1  -f  *)100  =  4769 ;    solve  by  taking   the   logarithm   of 
both  sides. 

15.  Discuss  the  increase  of  the  population  of  the  United 
States  from  1810  to  1910  as  in  problem  15  the  population  of 
New  York  City  is  discussed. 

3.  Graphical  representation  of  algebraic  functions. — To  repre- 
sent a  point  on  a  given  line  only  one  number  is  necessary  with 
a  point  of  reference  and  some  unit  of  length.  To  every  num- 
ber corresponds  one  point  and  only  one  and  .conversely  to 
every  point  corresponds  one  number  and  only  one  number. 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     67 


The  distance  cut  off  from  0,  the  origin,  by  any  point  on  this 
line  may  be  called  the  abscissa  of  the  point ;  a  moving  point 
upon  this  line  may  be  designated  by  the  variable  x,  which  is 
then  thought  of  as  assuming  different  values,  corresponding  to 
the  different  positions  of  the  point  upon  the  line. 


i       i       I       i       I       I       I 

_7     _e     -5     -4     -3     -2     -1 


I          I 


8        9 


If  another  scalar  line,  OF,  be  taken  intersecting  OX  at 
90°,  the  two  lines  may  be  conveniently  used  to  represent  the 
position  of  any  point  in  the  plane  of  the  two  lines.  The  two 
lines  of  reference  are  called  commonly  the  osaxis  and  the  y- 
axis  respectively. 

The  position  of  a  point  on  the  earth's  surface  is  given  by  a 
pair  of  numbers  representing  in  degrees  longitude  and  lati- 
tude ;  the  +  and  —  of  our  numbers  are  replaced  by  E.  and  W. 
in  longitude,  and  by  N.  and  S.  in  latitude.  If  we  agree  to  give 
longitude  first,  then  +  and  —  could,  in  both  terms,  replace 
the  letters,  and  position 
on  the  earth's  surface  of 
any  point  can  be  given 
by  a  pair  of  numbers. 
The  system  of  represent- 
ing points  in  a  plane  is 
not  essentially  different. 

Given  any  point  in  the 
plane  as  P,  a  perpen- 
dicular is  dropped  to  the 
horizontal  line.  The  dis- 
tance cut  off  on  this 
horizontal  line  is  called 
the  abscissa  or  cc-coordi- 
nate  of  P;  the  distance 
cut  off  on  the  vertical 
line  OF  by  a  perpen- 
dicular from  P  to  0  F  is  called  the  ordinate  or  ^-coordinate 
of  the  point  P  and  it  is  evidently  equal  to  the  _L  PM  dropped 


Location  of  points  in  a  plane 


68  UNIFIED  MATHEMATICS 

to  the  axis  OX.  The  two  numbers  together,  abscissa  given 
first,  serve  to  locate  the  point ;  thus  a  point  PI  1  units  to  the 
right  of  0  Y  and  5  units  above  OX  is  located  on  our  diagram. 
To  this  point  corresponds  the  pair  of  numbers  (7,  5)  (read 
"  seven,  five  ")  and  to  the  pair  of  numbers  (7,  5)  corresponds 
point  PI.  The  point  P4  symmetrical  to  PI  with  respect  to 
OX,  is  (7,  —  5)  the  negative  ordinate  indicating  that  the  point 
is  below  the  #-axis.  P2  (—  7,5)  and  P3  (—7,  —5)  are  located 
upon  the  diagram. 

A  moving  (or  a  variable)  point  P  in  the  plane  is  designated 
by  (x,  y"),  which  is  read  "x,  y"  (not  "x  AND  ?/"),  and  the 
coordinates,  abscissa  and  ordinate,  of  P  are  a  different  pair 
of  numbers  for  each  position  of  the  point,  i.e.  x  and  y  are 
variable. 

Every  point  represents  a  pair  of  numbers,  and  consequently 
a  series  of  points  will  represent  a  series  of  pairs  of  numbers. 
In  the  statistical  diagrams  the  pairs  of  numbers  are  numbers 
functionally  related.  In  an  algebraic  function,  y  —  3  x  +  5, 
we  have  involved  a  relationship  corresponding  to  a  mass  of 
statistical  information,  and  the  pairs  of  numbers  can  be  repre- 
sented upon  a  diagram  just  as  before.  Corresponding  num- 
bers, a  pair  of  numbers,  are  obtained  by  giving  a  value  to  x 
and  computing  the  value  of  y.  The  points  are  seen  to  lie 
upon  a  straight  line,  which  we  shall  see  includes  all  points  and 
only  those  points  whose  coordinates,  abscissa  and  ordinate, 
when  substituted  for  x  and  y,  respectively,  satisfy  our  given 
equation.  This  line  is  called  the  graph  of  the  function,  3  #+5, 
or  the  locus  of  the  equation,  y  =  3  x  -f-  5  ;  the  operation  of 
locating  the  points  and  connecting  them  is  termed  plotting 
the  graph. 

To  represent  on  cross-section  paper  any  equation  in  two 
variables  x  and  y,  t  and  s,  u  and  v,  or  by  whatever  letters 
designated,  two  intersecting  scales  as  axes  of  reference  OX 
and  OF,  OTand  OS,  or  OC/"and  OF  are  taken,  and  pairs  of 
values  which  satisfy  the  functional  relationship  are  plotted  as 
above. 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     69 

4.  Historical  note.  —  The  invention,  or  more  properly  the 
discovery,  of  analytical  geometry  was  made  in  the  early  part 
of  the  seventeenth  century.     The  first  work  directly  on  the 
subject  was  published  by  Rene  Descartes  in  1637,  La  G6om6- 
trie,  a  work  small  in  compass  but  great  in  its  effect  upon  the 
development  of  mathematics  and  science.     Almost  simultane- 
ously another  Frenchman,  Pierre  Fermat,  also  discovered  the 
methods  independently  of  Descartes. 

The  idea  of  coordinates,  called  Cartesian  after  Cartesius 
(Latin  form  of  Descartes)  was  not  new ;  in  fact,  as  we  have 
noted,  this  idea  is  found  in  the  latitude  and  longitude  of  Hip- 
parchus  (200  B.C.).  The  idea  of  coordinates  for  drawing  simi- 
lar figures  was  known  even  to  the  early  Egyptians,  and  this 
idea  was  used  for  surveying  purposes  by  Heron  of  Alexandria 
(c.  100  B.C.).  The  idea  of  fundamental  properties  of  any  curve 
as  related  to  its  axis  or  axes  or  to  tangent  lines  and  diameters 
was  also  not  new.  The  new  point  was  to  combine  these  ideas, 
referring  several  curves  and  straight  lines  to  axes  geomet- 
rically independent  of  the  curves,  using  letters  to  represent 
constant  and  variable  distances  associated  with  the  curves 
and  lines  involved ;  the  graphical  representation  of  negative 
quantities  is  a  vital  part  of  the  analytical  geometry.  These 
developments  were  made  both  by  Descartes  and  by  Fermat. 

Modern  mathematics  begins  with  this  analytical  geometry 
and  with  the  calculus  which  was  developed  within  a  century 
after  Descartes  by  Newton  and  probably  independently  by 
Leibniz. 

5.  Industrial  applications.  —  At  the  present  time  the  graphical 
representation  of  statistics  is  playing  an  increasingly  important 
role  in  many  industrial  enterprises.     Curves  derived  from  ob- 
servations, empirical  curves,  are  expressed  in  graphical  form 
for  convenience  of  reference  and,  frequently,  for  interpolation 
between  observed  values.     The  normal  distribution  curve  is 
employed  not  only  by  statisticians  but  also  in  the  production 
departments  in  many  factories  in  the  classification  of  their 
products. 


70 


UNIFIED  MATHEMATICS 


ILLUSTRATIVE  EXAMPLES 

To  plot  a  function  of  x,  give  x  values,  find  the  corresponding 
values  of  y,  or  conversely,  and  plot  the  points.  Connect  by  a 
smooth  curve  passing  through  all  the  points  in  succession 
moving  continuously  from  left  to  right. 

1.  Plot  the  graph  of 
the  function  of+ 4  x— 5, 
i.e.  plot  the  locus  of  the 
equation,  y=x* -\-kx-  5. 


Give  to  x  the  values  from 
0  to  3  and  from  0  to  -  6  ; 
beyond  these  values  in 
either  direction  the  values 
of  y  evidently  become  very 
large.  The  curve  is  evi- 
dently symmetrical  "with  re- 
spect to  a  line  parallel  to 
the  axis  and  2  units  to  the 
left. 

The  points  where  this 
curve  crosses  the  z-axis  rep- 
resent solutions  of  the  equa- 
tion x2  +  4  x  —  5  =  0. 


POINTS      POINTS 


Graph  of  y  —  x2  + 
2.    Plots  =  20  £+50. 


-5 


250 


200 


150 


100 


0  50 

1  70 
2|  90 
3110 
4130 
5150 
6170 
7190 
8210 
9230 

10  150 


1  70 
1.1  72 
1.274 
1.876 

1.478 
1.580 
1.6  82 
1.784 
1.886 
1.988 
290 


Upper  graph,  s  =  20 1  +  50  from  t  =  0  to  t  =  10,  upper  and  left-hand  scales 
Lower  graph,  s  =  20 1  +  50  from  t  =  1  to  t  —  2,  lower  and  right-hand  scales 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     71 


The  lower  and  right-hand  scales  would  be  used  if  you  were  interested 
in  the  behavior  of  the  function  in  the  interval  from  t  =  1  to  t  =  2.  By 
the  tenfold  enlargement  you  can  read  values  to  the  third  significant 
figure. 

This  may  represent  the  motion  of  a  body  which  starting  at  a  point 
50  feet  from  the  given  point  of  reference  moves  away  from  that  point 
in  a  straight  line  at  the  rate  of  20  feet  per  second.  The  units  might  be 
miles  and  hours,  so  that  the  speed  would  be  given  as  20  miles  per  hour  ; 
this  may  represent  then  the  motion  of  a  train. 

3.    Plot  y  =  a?  —  2&  —  18a;+24. 

The  values  of  y  are  so  large  that  the  figure  occupies  too  much  space 
vertically.  To  obviate  this  difficulty  one  square  on  the  axis  of  y  is  taken 
to  represent  ten  units  of  y  and  one  square  on  the  x-axis  is  taken  to 


-     lKi>44)   r 

|::iiE|;i|!|;i|j|y:j|E|;ii| 

POINTS 
ON  THE 

__             :::::::::::    CURVE 

<v  /                \        t  :  :  :  :  : 

:::  ::___::.._::__  :  ::::;:    x 
_5 

61 
0 
33 
44 
39 
24 
5 
-12 
-21 
+  16 
9 
48 

—  ->S"/—                       --  ^80  — 

:.:..      4 

g 

1  '/I/     I  I                                            V  3 

—  Y    -                                 \\- 

.    ':.:_./;:::|:_:::   ^  -1 

~f         —  •?        —  Y         —                                        \  \ 

l       :              /:>       t        "            0 

~/|::                                      --:sl 

%              .-T--'-                            i 

/                          10-      3 

^/ 

£c~  :     ~~:l                                  2 

'     7; 

,     / 

•>n  - 

:_.5t_,z:  :::::  ::.::  .::::  ::.::      3 

f 

(^           V'   'i                                                        /I 

/. 

.  c  u_ 

::;;:  fc:  ---  -30  1 

::::     5 

1J_LJ  

A 

Illl 

Graph  of  y  =  r»  -  2  x2  -  18  x  +  24 

represent  one  unit  of  x.  This  serves  to  compress  or  telescope  the  curve, 
but  the  essential  peculiarities  are  preserved.  In  particular  the  points 
at  which  the  curve  crosses  the  x-axis,  the  values  of  x  which  make 
x3  —  2x2—  18  x  +  24  =  0,  remain  unchanged.  These  values,  the  roots 
of  x3  —  2X2  —  18  x  +  24  =  0,  are  seen  to  be  —  4,  1.3,  approximately,  and 
4.8  approximately. 

In  general  an  algebraic  equation  of  this  type  is  not  likely  to  have  a 
rational  root,  such  as  the  —  4  above. 


72  UNIFIED  MATHEMATICS 

4.   Plot  a-2  +  y2  -  36  =  0. 


Graph  of  x-  +  y-  —  36  =  0 

In  drawing  the  graph  of  this  function  of  x  (implicit),  it  is  important 
to  note  that  there  are  two  values  of  y  corresponding  to  each  value  of  x, 
and  that  these  two  values  are  symmetrically  distributed  with  respect  to 
the  x-axis.  Similarly  this  curve  is  symmetrical  with  respect  to  the  ar-axis, 
since  any  value  of  y  gives  two  corresponding  values  of  x,  numerically 
equal  but  opposite  in  algebraic  sign.  The  points  when  located  are  con- 
nected by  a  smooth  curve  which  is  here  a  circle. 

To  this  diagram  reference  has  been  made  in  problem  7,  page  17.  As  a 
circle  of  radius  6  the  ordinates  at  x  =  1,  2,  3,  4,  and  5,  respectively,  give 
graphically  the  square  roots  of  35,  32,  27,  20,  and  11. 

The  more  complete  discussion  of  equations  of  this  type  is  given  in 
Chapter  XIV. 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     73 

5.   Plot  y  =  xz  for  a?  =  0,  .1,  .2,  •••,  1.0,  1.1,  1.2,  1.3. 

Note  that  precisely  the  same  curve  is  obtained  if  units  0,  1,  2,  3,  •••  15 
are  taken  instead  of  tenths  on  the  x-axis  and  tens  in  the  place  of  tenths 
on  the  vertical  axis,  as  indicated  on  the  lower  and  right-hand  scales. 


POINl 
THEC 
I.E 
HAND, 
UPI 
SCA 

X 
-.2 

1 

0 
.1 
.2 
.3 
.4 
.5 
.6 
.7 
.8 
.9 
1 

£ 

ir__ 

•8  ON 

THE  CURVE,  

RIGHT-                           _              II 

,       '  i 

/   : 

HAND,    AND 

ii;"i 

::£:_t-in- 

LOWER 

:  12::: 

SCALES      IIIIIIIIIII      :    ~~~~,~ 

nit  _ 

/  —  ±t±i 

v    :--- 

:     x 

y 

......  :x.-_i-oo- 

.04    - 

2 

4  : 

/              111' 

01    "  " 

i 

/             -  -H)0- 

0  :: 

0 

0    - 

r                  .\  \  \- 

m    S 

*-y—          ---80- 

.01    .1 

1 

i 

/ 

.04  ;i 

AQ 

2 

4  I::::::::::::::::::: 

.16   -Q 

4 

16    : 

iri:                 —  ^50- 

.25  : 

5 

25    "                                <s 

"T 

.36   g 

Kn 

6 

36                                     ' 

AQ 

Yl 

1-- 

(IA     . 

fi4.      -     -  -                           -4-  ^ 

. 

.81   ii 

9 

81    -                   / 

±t 

10 

—  ——  —M)- 

1   _± 

100   i-      iin;'ii:i;    :: 

-i 

1    I    I 

u 

~S 

_20- 

j-j-j- 

_.  <*"  __ 

1  1  | 

r  HI 

^/ 

..  +  ~1$~ 

^^ 

•v^. 

^-' 

:±:::S:::::::::::::X 

£    -\ 

"ijjr  '" 

2 

\    j"""  "3"    |1      Ij- 

,8       ,9     _t  0"  "  i"  1-  ~  " 

':  "     43 

_  _Li  LA  Li4- 

-  4i  -Hitttt5-  -nf  _  ^JL 

0 

+T 

^ffl^ffl^ffi 

Graph  of  y  =  x2 

This  line  if  drawn  somewhat  carefully  can  be  used  to  read  squares  of 
numbers  of  two  places  to  two  or  three  places.  Thus  (.85)2  =  .72  ;  (.96)2 
=  .92  ;  (.73)2  =  .54.  Square  roots  can  also  be  read  from  this  curve  by 
noting  the  horizontal  length,  corresponding  to  any  given  vertical  length. 
The  square  root  of  20  is  read  as  4.45,  of  30  as  5.50,  of  40  as  6.35,  of  50  as 
7.05,  of  60  as  7.75,  of  70  as  8.38,  of  80  as  8.95,  of  90  as  9.5 ;  the  square 
root  of  630  must  read  as  6.3  down  between  2  and  3  on  the  horizontal 
scale,  evidently  about  25. 


74 


UNIFIED  MATHEMATICS 


1.728 


Graph  of  y  =  x3,  using  three  different  scales 
The  portion  of  the  cubical  parabola, 


given  by  positive  values  of  x  from  0  to  10.  * 

The  cubes  of  numbers  from  0  to  10  can  be  read  quite  accurately  to 
two  significant  figures,  with  an  approach  to  the  third. 

Thus  (8.4)3  is  read  on  the  1  to  100  scale  curve  as  595,  instead  of  593  ; 
(4.4)3  is  read  on  the  1  to  10  scale  curve  as  850  instead  of  852  ;  (1.8)3  is 
read  on  the  1  to  1  scale  curve  as  6.00  instead  of  5.83. 


GRAPHICAL  REPRESENTATION  OF  FUNCTIONS     75 

PROBLEMS 

1.  Plot  y  =  y?,  from  x  =  \.  to  x  =  5,  using  one  half  inch  on 
the  vertical  axis  to  represent  10. 

2.  Plot  y  =  oc?,  from  x  =  1  to  x  =  10,  using  one  half  inch  on 
the  vertical  axis  to  represent  100.     See  graph,  above. 

3.  Plot  y  =  x?,  from  x  =  0  to  x  =  1  by  tenths,  using  10  half- 
inches  for  1  unit  on  each  axis. 

4.  Read  from  the  above  curves  the  following  cubes,  to  2 
significant  figures : 

3.23,  4.73,  .823,  1.53,  .643,  .583,  7.13,  9.23. 

5.  Plot  the  graph  of  s  =  200  t  —  16  t2,  from  t  =  1  to  t  =  12. 
This  represents  the  height  at  time  t  of  a  ball  thrown  upward 
with  a  velocity  of  200  feet  per  second. 

6.  Plot  the  graph  of  s  =  1000  -  200 1  —  16  t2,  from  t  =  1  to 
t  =  4 ;  this   represents  the  height  above  the   earth  of  a  ball 
thrown  down  from  the  top  of  the  Eiffel  tower  with  a  velocity 
of  200  feet  per  second. 

7.  Plot  the  graph  of  xz  +  y-  =  100  using  £  inch  as  1  unit  on 
each  axis.     Find  from  the  graph  ten  pairs  of  numbers  whose 
squares  summed  equal  100. 

8.  The  area  of  a  circle  is  given  by  the  formula  A  =  irr2 ; 
plot  the  graph  of  the  function  A  from  r  =  0  to  r  =  10  inches  ; 
use  3.14  for  ?r. 

9.  The  capacity  in  gallons  of  a  cylindrical  can  of  height  10 
inches  having  a  diameter  of  d  inches  is  given,  within  -J^  of 
1  % ,  by  the  formula  : 

C  =  -g-0-  •  «2  +  Ts-jnj-  "  "  • 

Plot  the  graph  for  d  =  1  to  20  and  interpret  as  gallons  per 
inch  of  height.  How  could  you  find  from  the  graph  the  ca- 
pacity of  a  can  having  a  diameter  of  8  inches  and  a  height  of 
9  inches  ?  Check  by  cubic  inches,  using  231  cubic  inches  to 
the  gallon. 


76  UNIFIED  MATHEMATICS 

10.    The  capacity  in  cubic  feet  of  a  silo,  d  feet  in  diameter, 
24  feet  high,  is  given  by  the  formula 

ir       TT(P  -24       a      n 

V=—  —  =67T<P. 

4 

Construct  a  graph  to  enable  you  to  read  the  capacity  of  silos 
having  diameters  varying  from  10  to  18  feet  and  give  approxi- 
mate accuracy  of  the  curve  in  the  neighborhood  of  d  =  11,  14, 
15,  and  17. 


CHAPTER   V 


1.  Theorem.  —  Any  equation  of  the  first  degree  in  two  vari- 
ables (x  and  y)  has  for  its  graph  a  straight  line. 

Proof.  —  The  general  equation  of  the  first  degree  may  be 
written  Ax  +  By  +  (7=0.  This  equation  can  always  be  put 
in  one  of  the  four  forms: 


x  =  k,    if  B  =  0,  or  if  B  =  0  and  (7=  0  ; 
y  =  k,    if  .4  =  0,  or  if  .4  =  0  and  (7=0; 
v  =  mo;,  if  (7=0, 

H-.,^ 

line   parallel  to   the  y-axis,  at        J-^.'- 
a  distance  k  units  from  it;  on 
such  a  line  the  abscissa  of  any     T  Hfi  $E 

_  _   [  [  «v~  IS3I 

point  is  constant.     The  coordi- 

4—  -P  

FU-;-;!;;:;!:-;; 

t   . 

TTTT   ""' 

j  ?  •  T 

•  r  S      H 

nates  of  any  point  on  the  line     — 

:!  T|^§fti 

;         -  I  ; 

satisfy  the  equation,  and  any     p5    -^ 

..1       L'    : 

:;       A 

point  whose  coordinates  satisfy 
Thus,  x  —  —2  or  x  +  2—0  repre- 

-  -1—  r 

—  
_  -__  

and  2  units  to  the  left  of  the 
y-axis.     Similarly,  y  —  k  repre-     ill 
sents  a  straight  line  parallel  to 

the  avaxis  and  at  a  distance  of 
(_ 

k  units  from  it.                                 and       G 

77 

rraph  of  x  +  2  =  0 
rraph  of  x  —  k  =  0 

78 


UNIFIED  MATHEMATICS 


y  =  mx.  Assume  different  points  which  satisfy  this  rela- 
tion ;  the  origin  lies  upon  the  locus  ;  (xl}  yj  and  (x2,  y2)  satisfy, 

if  #!  =  mxi,  y2  =  mx^. 

Any  two  points  (x1}  y,),  (a*,  y») 
which  satisfy  this  equation  can  be 
shown  to  lie  upon  the  straight  line 
connecting  either  one  with  the 
origin,  which  evidently  satisfies 
the  equation. 

Consider  first  m  to  be  positive ; 
the  point  (xl,  y^)  may  be  taken  in 
the  first  quadrant. 

Since  yl  =  mxl}  and  y2  =  wio;2, 


—  =  —  =  m. 

X!         X2 

The  right  triangle  containing  ccj 
and  yi  as  the  sides  is  similar  to 
the  right  triangle  with  x2  and  y2 

as    sides,    since    £isa2l«      Hence 


Graph  of  y  =  mx 


the   corresponding    angles   at    0  are   equal   and    the    points 
(x1}  2/j)  and  (x2,  y^  lie  upon  a  straight  line  through  the  origin. 

NOTE.  —  If  yz  and  Xz  are  negative,  — 2  is  in  truth  to  be  replaced  by  ~  ™ 
since  only  positive  quantities  are  involved  in  plane  geometry. 

Conversely  any  point  (x,  y)  which  lies  upon  the  given  line 
drawn  satisfies  the  given  equation.     For,  by  similar  triangles, 

-  =  —  =  m,  whence  y  —  mx. 

TV 
JU  A/1 

When,  secondly,  m  is  negative,  the  argument   is  slightly 
changed,  since  any  point  (xi}  y^  which  satisfies  the  equation 

must  have  coordinates  opposite  in  sign;  then  either    ^'     or 
Vi 


i 


, 

equals  —  m. 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        79 

In  the  equation  y  =  mx  +  k  for  any  values  of  x,  the  corre- 
sponding values  of  y  are  greater  by  k  than  the  corresponding 
ordinates  of  y  =  mx.  Construct  any  three  such  ordinates  of 
y  =  mx  -f  k.  Since  these  exten- 
sions are  parallel  and  equal  in 
length,  parallelograms  are  formed  ; 
the  inclined  sides  of  these  paral- 
lelograms are  parallel  to  the  line 
y  =  mx  and  consequently  to  each 
other.  Since  any  two  of  these 
parallel  lines  have  a  point  in  com- 
mon, they  coincide  and  form  one 
straight  line  (by  plane  geometry). 
The  value  m  is  called  the  slope  of 
the  line  y  =  mx  +  k,  and  evidently 
varies  as  the  angle  which  the  line 
makes  with  the  #-axis  varies.  The 
angle  which  a  line  makes  with  the 
o-axis  is  termed  the  slope  angle  of 
the  line. 

Conclusion.  —  Since  every  equa- 
tion of  the  first  degree, 

Ax  +  By  +  C  =  0, 
can  be  put  into  one  of  four  forms,  mentioned,  every  equation 
of  the  first  degree  represents  a  straight  line. 

Conversely,  every  straight  line  is  represented  by  an  equa- 
tion of  the  first  degree ;  if  the  line  is  parallel  to  one  of  the 
axes,  the  form  of  the  equation  is  evidently  x  =  k  or  y  —  k ; 
if  the  line  passes  through  the  origin,  the  form  is  y  =  mx;  and 
every  point  on  the  line  can  be  shown  to  satisfy  this  equation, 
for  let  (xl)  2/j)  be  any  fixed  point  on  the  line  and  (x,  y)  any 
point  whatever  on  the  line,  then 

21  =  ^ ,  by  similar  triangles,  whence 


tJ 

-  =  m,  a  fixed  value,  and  y  =  mx. 


80 


UNIFIED  MATHEMATICS 


Any  other  line  will  be  parallel  to  a  line  through  the  origin, 
and  its  points  will  satisfy  an  equation  of  the  form 

y  =  mx  +  k. 

On  the  line  y  =  mx  +  k,  if  at  any  point  the  value  of  x  is 
increased  by  one  unit,  the  value  of  y,  the  function  of  x,  is 
increased  by  m  units ;  on  this  line  everywhere  y  increases  m 
times  as  fast  as  x;  the  ratio  of  the  increase  of  y  to  the 
unit  increase  of  x,  m,  gives  on  the  straight  line  the  rate 
of  change  of  the  function  y  as  compared  with  the  rate  of 
change  of  x. 

2.    Intersection  of  graphs.  — 

Plot  2  x  +  3  y  -  26  =  0, 

x  +  y  —  5  =  0. 

Every  point  on  the  first  line  is  such  that  its  coordinates 
(x,  y}  when  substituted  in  2x  -\-  3y  —  26  =  0,  satisfy  the 

equation;  there  are  an 
infinite  number  of  such 
points,.  e.0.(0,  V)>  (1,  8), 
(2,  ^),  (13,  0),  (-'/,  1), 
(-1, -238-),  ~.  By  substi- 
tuting 0  or  1  or  2  or  —  1, 
—  2,  •••,  for  x  and  solving 
for  y,  or  conversely,  points 
are  obtained  whose  coor- 
dinates satisfy  the  given 
equation ;  similarly  every 
point  on  the  second  line 
is  such  that  its  coordinates 
satisfy  the  equation, 


Graphs 

-^          y          w  —  w 

the  point  of  intersection 

satisfies  both  equations,  and  its  coordinates  can  be  obtained  by 
solving  the  two  equations  as  simultaneous.     The  argument  is 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        81 

entirely  similar  for  the  points  of  intersection  of  any  two  loci, 
representing  algebraic  equations  ;  the  points  of  intersection 
satisfy  both  equations,  and  give  a  graphical  method  of  ap- 
proximating the  solutions  of  the  equations  regarded  as  simul- 
taneous. 

To  review  this  demonstration,  answer  the  questions  below, 
and  read  the  discussion. 

What  is  true  concerning  the  coordinates  of  every  point  on 
the  first  line  ?  on  the  second  line  ?  What  is  true  concerning 
the  point  of  intersection  so  far  as  the  two  given  equations  are 
concerned?  The  drawing  shows  that  (—  11,  16)  satisfies  both 
the  equations,  and  substitution  shows  that  this  is  precisely 
correct.  In  general  the  graphical  solution  is  only  approxi- 
mate, the  degree  of  accuracy  depending  upon  the  accuracy  of 
the  drawing  and  the  scale  used. 

The  point  of  intersection  of  two  straight  lines  represents  graph- 
ically the  solution  obtained  by  solving  the  two  equations  as 
simultaneous. 


Graphs 


Intersections  of  y  =  3  x,  and  x*  +  y2  =  25. 

The   graphical   presentation  shows  very   plainly   that   the 
solution  is,  approximately, 

a;  =  1.6 
and  y  =  4.8. 


82 


UNIFIED  MATHEMATICS 


-1-0 


Intersections  of  y  =  x2,  y  =  3  x  +  5. 

The  graph  shows  that  there  are  two  solutions  ;  in  the  one, 

a?  =  -1.2,    y  =  +  1.4, 
and  in  the  other,        a;  =  4.2,        y  =  17.8. 

These  are  approximate  values. 

Plot  carefully  the  graphs  of  the  preceding  problems,  check- 
ing on  the  work  presented  by  the  graphs. 

Plot  carefully  these  two  lines  and  verify  the  statements 
made: 

2  x  +  3  y  —  26  =  0,  |   Graphically,  parallel ; 
2x-\-3y  —    8  =  0.  j  algebraically,  no  solution. 

The  point  of  intersection  of  two  graphs  represents  graphically 
the  solution  of  the  two  equations  regarded  as  simultaneous. 

3.  Intercepts.  —  Any  given  line  or  curve  cuts  off  on  the  co- 
ordinate axes  distances  that  are  called  the  intercepts  of  the 
line  or  curve.  The  ^intercept  is  obtained  analytically  by  sub- 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        83 

stituting  y  =  0  and  solving,  i.e.  by  solving  as  simultaneous  the 
equations  of  the  £-axis  and  the  given  line ;  the  ^-intercept  is 
obtained  by  substituting  x  =  0. 

The  ^-intercept  of  2  x  +  3  y  —  26  =  0  is  13,  obtaining  by 
substituting  y  =  0  in  2  x  +  3y  —  26  =  0;  the  ^-intercept  is 
+  -2^6 ;  of  a2  +  y2  =  25,  the  cc-intercepts  are  ±  5,  the  y-inter- 
cepts  are  also  ±  5. 

Note  that  the  problem  of  finding  the  intercepts  of  a  given 
graph  is  a  special  case  of  the  problem  to  find  the  intersections 
of  two  given  curves ;  the  OS-intercept  designates  the  inter- 
section of  the  given  curve  with  the  a-axis,  y  =  0,  and  similarly 
the  ^-intercept  refers  to  the  intersection  with  x  =  0. 

Rule.  —  To  find  the  ^intercept,  put  y  =  0,  and  solve ; 
similarly  for  the  ^-intercept. 

4.  Pencil  of  lines.  —  The  straight  lines  which  pass  through 
a  common  point  constitute  what  is  termed  a  pencil  of  lines. 
If  the  common  point  is  determined  as  the  intersection  of  two 
given  lines,  we  may  write  the  equation  of  the  pencil  of  lines 
in  terms  of  the  two  expressions  which  put  equal  to  zero  repre- 
sent the  given  lines. 

The  pencil  of  lines  through  the  intersection  of 

y-3x-5=0  (IJ 

3y  +  2aj  +  7  =  0  (72) 

is  given  by  the  linear  equation,  A;  being  assumed  constant, 
(3)  y-3x-5  +  k(3y  +  2x  +  7)  =  0.  (/3) 

Evidently  any  point  on  the  first  line,  ?„  makes  y  —  3x  —  5  =  0, 
and  any  point  on  the  second  line,  Z2,  makes  3y-{-2x-\-7  =  0; 
the  point  of  intersection  substituted  in  our  equation  (3)  gives 
0  +  k  •  0  or  0,  hence  the  point  of  intersection  of  lt  and  12 
satisfies  equation  (3)  for  all  values  of  k. 

By  giving  k  successive  values  13  can  be  made  to  pass  through 
any  point  of  the  plane.  Thus  to  pass  through  (1,  5)  sub- 
stitute (1,  5)  in  13  and  solve  for  k,  giving 

5  _  3  -  5  +  k(15  +2  +  7)  =  0, 


84  UNIFIED  MATHEMATICS 

or  24  k  =  3,  fc  =  £.  The  line  y  —  3ar  —  5  +  |(3y+2*  +  7)  =  0, 
or  8#-24  a-40+  3  y  +2  x+7  =  0  reduces  to  lly-22x—  33=0, 
or  y  —  2  x  —  3  =  0  when  simplified. 

In  solving  as  simultaneous  the  two  equations  y  —  3x  —  5  =  0, 
and  3y  +  2a;  +  7  =  0,  the  particular  lines  parallel  to  the  axes 
of  reference  and  passing  through  the  point  of  intersection  of 
/!  and  Z2  are  sought.  Thus,  after  multiplying  the  upper 
expression  by  —  3  and  adding,  you  get  the  line  /3  with  k  =  —  ^. 


gives  —  11  x  —  22  =  0,  or  x  =  —  2. 

To  eliminate  x  we  multiply  the  upper  expression  by  2  and 
the  lower  by  3  and  add  ;  this  gives,  11  y  +  11  =  0,  or  y  =  —  1. 
The  same  line  given  by  11  y  +  11  =  0  is  obtained  from  line 
1$  with  k  =  |  ;  i.e.  : 
y-3x-5  +  %(3y  +  2x  +  7)  =  0 

gives  11  y  +  11  =  0,  or  y  +  1  =  0. 

The  point  of  intersection  of  the  two  lines,  (—2,  —  1),  is 
given  as  the  intersection  of  x  =  —  2  and  y  =  —  1. 

PROBLEMS 

1.  Solve 

y-3x  —  5  =  0, 
3y  +  2x—  7  =  0, 
both  graphically  and  algebraically. 

2.  Plot  the  graphs  of 


x  +  y  —  2  =  0. 

Do  these  three  lines  appear  to  meet  in  one  point  on  your 
diagram?  Have  these  three  equations  a  common  solution? 
Substitute  the  solution  of  the  first  pair  (obtained  in  problem  1) 
in  the  third  equation.  Later  it  will  be  shown  that  a  point 
whose  coordinates  when  substituted  in  a  first-degree  expression 
give  a  small  numerical  value  is  near  the  straight  line  repre- 


THE  LINEAR  AND   QUADRATIC  FUNCTIONS        85 

sented  by  the  equation  formed  by  putting  that  expression  equal 
to  zero. 

3.   Plot  15  points  whose  coordinates  satisfy  the  equation 


4.  Plot  the  lines  x  =  3  and   y  =  4  ;    what  point   is   repre- 
sented by  these  equations?     Note  that  the  Cartesian  system 
(x,  y)  of  representing  points  implies  each  point  as  the  inter- 
section of  two  lines. 

5.  Solve 


both  graphically  and  algebraically. 

6.  The  weight  of  a  cylindrical  vessel  of  water  when  filled 
to  a  height  of  10  inches  is  6.8  pounds,  when  filled  to  a  height 
of  6  inches  it  is  4.4  pounds  ;  plot  the  two  points  (6,  4.4)  and 
(10,  6.8).    The  straight  line  joining  these  two  points  gives  the 
weight  of  the  vessel  when  filled  to  any  height  from  0  to  10. 
The  equation  may  be  written  iv  =  k  •  h  -f  c,  where  w  and  h  are 
the  variable  weight  and  height,  k  and  c  are  constants.     This 
equation  is  the  simple  statement  of  the  fact  that  the  weight  of 
the  water  and  the  container  for  any  height  h  is  the  weight  of 
the  vessel  (c)  plus  h,  the  height,  times  the  weight  of  the  water 
which  fills  the  container  to  a  height  of  one  inch.     Note  the  sig- 
nificance of  the  intercepts. 

pn  \ 

7.  The   equation,   w  =    ^   v,  may  be  used  to  express  the 

relation  between  the  volume  in  cubic  inches  and  the  weight  in 
pounds  of  a  given  mass  of  water..  Plot  this  carefully  and  find 
approximately  the  weights  of  100  cubic  inches,  500  cubic  inches, 
and  700  cubic  inches  of  water.  Find  the  volume  of  15  pounds 
of  water  ;  the  volume  of  25  pounds  ;  of  30  pounds. 

8.  The  volume  of  mercury  at  any  temperature  between  0 
and  40°  C.  is  given  by  the  equation    V=  fc(l  -f-  at),   wherein 
a  =  .00018  ;  for  A;  =1000  cu.  cm.  this  becomes  V=  1000  +  .18  t. 


86  UNIFIED  MATHEMATICS 

Plot  this  equation  taking  the  horizontal  axis  as  at  1000.  This 
is  equivalent  to  plotting  the  increase  in  volume,  /=.18<. 
Plot  for  0  to  40°  C.  and  find  the  increase  in  volume  when  1000 
cu.  cm.  of  mercury  at  0°  C.  are  heated  to  27°  C. 

9.   Find  the  equation  of  the  straight  line  through  (—3,  5) 
and  through  the  intersection  of  3x  —  y  —  7  =  0  and 

5  x  +  12  y-  17=0. 

10.  Plot  degrees  Fahrenheit  as  abscissas  and  degrees  Centi- 
grade as  ordinates,  connecting  (32°  F.,  0°  C.)  to  (212°  F.,  100°  C.), 
by  a  straight  line.     Find  the  equation  of  this  straight  line. 
Find  the  Centigrade  reading  corresponding  to  0°  Fahrenheit, 
to  100°  F.     Discuss  the  meaning  of  the  slope  of  the  line. 

11.  Find  the  intercepts  of  the  line  9  y  —  5  x  =  —  160.     Com- 
pare with  your  result  in  the  preceding  problem. 

12.  Plot  the  graph  of  s  =  16  tz,  for  values  of  t  from  t  =  0  to 
t  =  5,  using  one  inch  for  1  second  on  the  horizontal  axis,  and 
1  inch  for  100  feet  on  the  vertical  axis.     Find  value  of  s  when 
t  =  4.3  from  the  graph.     Check  by  computation. 

13.  Plot  carefully  x2  +yz  =  64,  and  y  =  3  x  —  5.     From  the 
graph  get  the  approximate  solution. 

14.  Show  graphically  how  to  change  a  system  of  marks  from 
a  scale  of  100  to  a  scale  of  75 ;  from  75  to  100. 

15.  Sound  travels  at  the  rate  of  1089  feet  per  second  in  air 
at  32°  F.  (or  0°  C.)  ;  at  the  rate  of  1130  feet  per  second  in  air 

13  t 
at  70°  F.     The  formula,  v  =  1054  +^  gives  very  closely  the 

\a 

velocity  in  feet  per  second  at  temperature  t°  Fahrenheit. 
Plot  the  graph  of  the  function,  plotting  the  excess  above  1000 
feet  as  ordinates  and  temperature  Fahrenheit  up  to  80°  F.  as 
abscissas.  At  what  temperature  is  the  velocity  1100  feet  per 
second?  How  would  you  adapt  these  figures  to  the  Centigrade 
scale  for  temperature  beginning  0°  C.?  v  =  1089  +  2  t  is  the 
resulting  equation. 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        87 

16.  The  velocity  after  t  seconds  of  a  bullet  shot  straight 
upwards   at  800   feet   per   second  is  given  by  the  equation 
v  =  800  -  32  t.     Plot  the  graph,  taking   100   feet  as   |   inch 
on  the  vertical  axis,  and  5  seconds  as  •£  inch  on  the  horizontal 
axis  ;  negative  values  of  v  mean  that  the  bullet  is  descending. 

17.  Plot  v  =  600  -f  32  1,  and  interpret  as  downward  velocity 
of  an  object  thrown  downwards  from  a  height. 

18.  Time  yourself  on  plotting  the  following  10  lines  ;  five 
may  be  plotted  with  respect  to  one  set  of  axes  : 

a. 


__ 

b.  3y  =  2x-5.  '    3     5 

c.  X-y-S  =  Q.  h.     </  =  fa 

d.  ' 


f. 

19.  Time  yourself  (a)  on  finding  the  slope  of  each  of  the 
ten   lines  in  problem  18  ;    (6)  on   finding    the   cc-intercept   of 
each  line  ;   (c)  on  finding   the   ordinate   of   the  point   whose 
abscissa  is  2  ;  (d)  on  finding  to  one  decimal  place  the  ordinate 
of  the  point  on  each  line  whose  abscissa  is  2.4  ;  (e)  on  putting 
these  lines  in  normal  form. 

20.  Plot,  using  values  correct  to  1  decimal  place,  the  follow- 
ing lines  : 

a.   3.1  a  +  4.5  0-12  =  0. 
6.   3.2  y  =  2.6  a  -5.7. 
c.   .9  a?  -4.8  ^8.3  =  0. 

5.   The  quadratic  function  of  one  variable.  —  Any  equation  of 
the  form  ax  +  b  =  0  is  called  a  linear,  or  first-degree  equa- 

tion, in  the  variable  x  ;  the  solution  is  given  by  x  =  --  ;  the 

a 

graph  of  the  function,  y  =  ax  4-  b,  is  a  straight  line  of  slope  a, 
with  the  y  intercept  equal  to  6,  and  with  the  x  intercept  repre- 
senting the  solution  of  the  equation,  ax  +  b  =  0. 

Any  equation  of  the  form  ax*  -f-  bx  +  c  =  0  is  called  a  quad- 


88  UNIFIED  MATHEMATICS 

ratio  equation  in  x;  a,  6,  and  c  are  to  be  regarded  as  con- 
stants. The  graphical  solution  of  one  equation  of  this  type 
has  been  presented  (page  70)  and  the  algebraic  solution  is 
given  in  elementary  algebra,  but  will  be  given  here  a  rapid 
review. 

Algebraical  solution  of2x2  +  8x-}-7  =  0  and  of  the  general 
equation,  ax*  +  bx  +  c  =  0 ; 

2  x2  +  8  x  +  1  =  0,  ax2  +  bx  +  c  =  0, 

x  =  -f,  a;2+  — =  --, 

a  a 


2  ay       4  a2      a 
b2  —  4  ac 


x  +  2  =  ±  .71  (or  .707  to  3  places),         a  +  —  =  ±  -~ 

2a  2  a 


O   r-1  —   &    ±  V&2  —  4  OC 

a;  =  —  1.29  or  —2.71,  x  =  —  —  . 


The  necessary  third  term  to  complete  the  square  is  obtained 
by  comparison  with  (x  ±  A:)2  =  x2  ±  2  kx  +  A-2. 

Graphically  the  equation  y  =  2x2  -}-  8x  -(-7  represents  a  curve 
which  intersects  the  a>-axis,  y  =  0,  in  the  two  points  whose 
abscissas  satisfy  the  equation,  2  #2-f-  8  #+  7=0.  y=2.T2+8#+8 
represents  a  curve  which  is  tangent  to  the  o^axis,  correspond- 
ing to  the  fact  that  the  roots  of  the  equation,  2x*  +  8o;  +  8  =  0, 
are  equal  to  each  other.  The  equation  y  =  2a^  +  8x  +  ll  repre- 
sents a  curve  which  does  not  cut  the  cc-axis,  corresponding  to 
the  fact  that  the  quadratic  2x2  +  8x-|-ll  =  0  has  for  solu- 


tions, x  =  —  — ,  values  corresponding  to  no  points  on 

Z 

the  x-axis,  i.  e.,  to  imaginary  values  of  x.     Plot  the  graphs  in- 
dicated. 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        89 

The  quadratic  equation  is  solved  algebraically  by  reducing 
the  problem  to  the  solution  of  two  first-degree  equations : 

b        -f-  V&2  —  4  ac 
2~a~  2a       ' 


. 
2a  2a 

The  quantity  &2  —  4  ac  which  appears  under  the  radical  sign  is 
called  the  discriminant  of  the  quadratic.  The  nature  of  the 
roots  of  the  quadratic  equation  is  determined  by  this  dis- 
criminant, when  a,  b,  c  represent  real  quantities,  i.e.,  a,  b,  and 
c  having  values  which  can  be  represented  by  points  upon  a 
scalar  line.  When 

b2  —  4  ac  >  0,  i.e.  positive,  the  two  roots  are  real  and  un- 
equal, when 

62  —  4  ac  =  0,  the  roots  are  real  and  equal,  and  when 
b-  —  4  ac  <  0,  i.e.  negative,  the  roots  are  imaginary. 

Further,  the  condition  that  the  roots  of  the  quadratic  should  be 
equal  given  by  b-  —  4  ac  =  0,  may  be  obtained  by  inspection, 
or  by  actually  setting  the  two  roots  equal  to  each  other  and 

simplifying;   ax2+  bx  +  c   may   then  be  written  a(x+- — )• 

V        2  a,/ 

Graphically  these  conditions  correspond  to  the  fact  that  the 
curve  y  =  ax2  +  bx  +  c  cuts  the  co-axis  in  two  points,  is  tan- 
gent to  the  ;K-axis,  or  does  not  intersect  it  at  all,  according  as 
b2  —  4  ac  is  greater  than,  equal  to,  or  less  than  0. 

Frequently  the  two  roots  of  the  quadratic  ax2  +  bx  +  c  =  0 
are  designated  by  xt  and  x2. 

Thus  x,  =  -6+V62 

2a 


,  -  b  —  V62  —  4  ac 

and  x2  = . 

2a 

The  sum  and  the  product  of  the  roots,  xl  +  x2  and  XiX2,  are 

b  c 

given,  respectively,  by and  +  - .     The  expressions  o^  -f  x.2 

a  a 

and  x&z  are  representative  symmetric  functions  of  the  roots  of 


90  UNIFIED   MATHEMATICS 

a  quadratic  function  of  one  variable,  being  expressions  which  re- 
main unchanged  when  a^  and  x2  are  interchanged. 

6.  Historical  note.  —  The  solution   of  linear   equations  was 
known   four  thousand  years  ago  to  ancient  Egyptians.     The 

equation  x  +  -  =  19,  was  proposed  and  solved  in  the  work  of  an 

Egyptian  writer  named  Ahrues ;  the  problem  reads,  with  "  ahau  " 
representing  "  heap  "  or  "  unknown,"  "  ahau  and  its  seventh, 
it  makes  19."  In  other  ancient  Egyptian  documents  problems 
leading  to  pure  quadratics  are  found.  The  Greeks  were  able 
to  give  as  early  as  450  B.C.  a  geometrical  solution  of  any  quad- 
ratic having  positive  roots ;  the  numerical  application  appears 
in  Greece  somewhat  later.  In  India  numerical  quadratics 
were  solved  in  the  fifth  and  sixth  centuries  A.D.  The  first 
systematic  treatise  combining  clearly  analytical  statement  with 
geometrical  illustration  is  given  by  an  Arab,  Mohammed  ibn 
Musa  al-Khowarizmi,  about  825  A.D.  His  work  continued  in 
use  for  centuries.  The  complete  quadratic  with  general, 
literal  coefficients,  did  not  come,  of  course,  until  after  the  in- 
troduction of  literal  coefficients  by  Viete  late  in  the  sixteenth 
century. 

7.  Graphical  solution  of  the  general  quadratic  equation.  —  The 
general   quadratic   equation  ax2  +  bx  +  c  =  0  can  be   solved 
graphically  by  means  of  one  fixed  curved  line,  y  =x2,  and  a 
variable  straight  line.     The  intersection  of 

y  =  x* 
and  ay  -f  bx  +  c  =  0 

gives  the  solution  of  the  equation  ax*  +  bx  +  c  =  0,  for  the 
solution  is  obtained  algebraically  by  substituting  for  y  its 
value  x2  in  ay  +  bx  +  c  =  0,  giving  ax2  -j-  bx  4-  c  =  0. 

The  graphical  solution  of  the  quadratics,  2  #2  —  6x—  5  =  0, 
2xz-6x  =  0,  2xz-  6x  +  f  =  0,  and  2x2—  6x+  10  =  0, 
is  presented  upon  the  diagram ;  the  student  is  urged  to 
solve  these  equations  algebraically  and  to  trace  the  correspond- 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        91 


Graphical  solution  of  quadratics 
2x2  — 6x-5  =  0;  2x2—  6x  =  0;   2x2-6x  +  f  =  0;  2x2-6x  +  10  =  0. 


Two  real  solutions. 


2y-6z  +  f  =  0; 
Two  coincident  solutions. 


Two  real  solutions. 


4y-  6x  +10  =  0. 

Two  imaginary  solutions. 


92  UNIFIED  MATHEMATICS 

ence  between  the  algebraic  and  graphical  solutions.  Two 
sets  of  real  and  different  roots  are  indicated  by  two  of  these 
straight  lines  on  our  diagram ;  one  set  of  equal  roots  is  in- 
dicated ;  one  pair  of  imaginary  roots  is  indicated  by  the  line 
which  does  not  meet  the  curve. 

The  graphs  of  the  corresponding  functions, y  =  2xz  —  6x  —  5, 
y  =  2  &  —  6  x,  etc.  should  also  be  drawn. 

PROBLEMS 

1.  Plot  the  graph  of  y  =  3x—  7;    give   to  x  the  integral 
values  from  —  2  to  -f-  5  and  find  the  values  corresponding  of  y. 

2.  A  freely  falling  body  falls  from  rest  in  t  seconds  a  dis- 
tance s,  given  by  s  =  16 12 ;  plot  points  given  by  corresponding 
values,  using  horizontal  axis  as  £-axis,  and  vertical  axis  for 
distance.     Take  values  of  t  from  0  to  10,  and  as  s  will  vary 
from  0  to  1600  take  1  cm.  to  represent  100  on  the  s-axis. 

3.  The  simple  interest  on  $  100  for  n  years  at  5  %  is  given 
by  1=  5n ;  plot  n  on  the  horizontal  axis  and  /  on  the  vertical. 
On  the  same  axes  plot  A  =  100  +  5  n  where  A  is  the  amount 
at  the  end  of  n  years,  plotting  A  on  the  vertical  axis.     On  the 
same   axes  plot  A=  100(1  -f  .05)"   for   «  =  1   to   10,   finding 
the  values  of  (1.05)2,  (1.05)3  ...  by  logarithms ;  this  gives  the 
amount  at  compound  interest,  compounded  annually.     Check 
by  the  table  at  the  end  of  the  book. 

4.  Take  problem  3,  using  4  %  as  the  interest  rate  and  6  % 
as  the  interest  rate  ;  take  the  values  of  (1.04)n  and  (1.06)"  from 
the  tables. 

5.  Take  the  data  of  problem  3,  using  6  %  simple  interest 
paid    semiannually,    and    6  °f0    interest    compounded    semi- 
annually. 

6.  The  velocity  of  a  freely  falling  body  is  given  by  the 
formula,  v  =  32 1,  when   falling   from   rest ;    or  v  =  32  t  +  k, 
where  k  represents  the  velocity  at  the  instant  when  t  =  0,  or  k 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        93 

is  the  velocity  at  the  instant  when  you  begin  to  measure  the 
time.  Plot  for  values  of  t  from  0  to  10. 

7.  A  bullet  shot  straight  up  into  the  air  at  a  velocity  of 
1000  feet  per  second,  has  its  height  above  the  earth  given  by 
the  equation  h  =  1000 1  —  16 1*.     Plot  this  equation  for  values 
of  t  increasing  by  intervals  of  5  seconds  from  t  =  0  to  t  =  100. 
If  the  bullet  is  shot  at  an  angle  in  such  a  way  that  the  vertical 
velocity  when  leaving  the  gun  is  1000  feet  per  second,  the 
given  equation  continues  to  hold  for  the  height  of  the  bullet 
above  the  earth.     The  resistance  of  the  air  (considerable  at  the 
velocity  mentioned)  is  neglected  in  these  equations. 

NOTE  ox  NUMERICAL  APPROXIMATIONS. — In  several  of  the  problems 
below,  as  well  as  in  some  of  the  preceding  problems  like  the  fifteenth  of 
the  preceding  set,  numerical  approximations  are  given  involving  simple 
fractional  expressions  as  substitutes  for  decimal  values.  The  method  of 
making  this  kind  of  substitution  which  is  frequently  of  use  involves 
simply  the  application  of  addition  and  subtraction  to  aliquot  parts  of  100, 
rejecting,  with  discretion,  places  which  are  not  necessary  to  attain  the 
degree  of  accuracy  warranted  by  the  data  of  the  problem. 

|  =  .33333,  |  of  ^  =  .033333,  etc. 

I  =  .25,  J  of  ^  =  .025,  etc. 

£  =  .16666,  I  of  ^  =  .016666,  etc. 

j  =  .50,  Jof  j^=.05,  etc. 

|  =  .125,  |  =  .875,  J  of  any  number  is  usually  obtained  by  subtracting 
J  of  the  number  from  the  number. 

\  =  .666666  ;  f  =  .75. 

Thus,  .365  =  |  +  ^  -  ,$ff  ; 

.1918  =  J-  +  jfr  +  T^nr  5  the  error  &  about  'As  of  *  % 
.1492  =  ^  +  A  —  TtfW  5  tne  error  is  at)0ut  }  of  1  %. 

These  numbers  are  chosen  at  random. 

8.  The  volume  in  gallons  of  a  cylindrical  container,  meas- 
ured in  inches,  is  given  by  V  =  —     ^— .      Replace   -     7r '     • 

4  X  AOL  4  X  2ioL 

by  its  approximate  value,  .003399,  or  \  of  1  %  of  cPh  +2% 
of  this  result,  or  (1.02)  x  ^  %  of  cPh ;  i.e.  use  the  formula 
F=1.02  x(.00£)d2/i.  Plot  values  from  d  =  1  to  d  =  20,  for 
h  =10. 


94  UNIFIED  MATHEMATICS 

9.    The  volume  in  barrels  per  foot  of  height  of  a  cylindrical 

cistern  is  given  by  V=  —  —  X  —  —  ,  in  which  7.48  is  the  number 
4       31.5 

of  gallons  per  cubic  foot  and  31.5  is  the  number  of  gallons  per 

barrel  ;  d  is  to  be  measured  in  feet.     Use  for   -  -  -^—  -    the 

4  x  31.5 

value        +     A  plotting  then   F  =  a+Tfo)d*,  for  d=l  to 


10.    Plot  the  number  of  barrels   per   foot  of   height  of  a 
square  tank  using  the  formula  F  =  —    —^  —  ,  computing  the 


7  48 

quotient   -      -  by  logarithms   and   using  this    multiplier  for 
31.5 

values  of  d  from  1  to  20.     Choose  appropriate  scale  to  get  the 
data  on  paper. 

11.  Given    h  =  800  1  —  16  P,    find    t    when    h  =  100,    1000, 
10,000,    12,000    respectively.      This   equation   represents   the 
height  to  which  a  bullet  would  rise  when  shot  vertically  up- 
wards at  a  velocity  of  800  feet  per  second,  neglecting  air- 
resistance.     Interpret  your  results.     This  bullet  has  a  velocity 
at  time   t,  v  =  800  —  32  1  ;   find  the  velocity  at   the   various 
heights  mentioned. 

12.  Solve  16  1-  —  800  1  +  7i  =  0  for  t,  regarding  h  as  a  con- 
stant.    See  preceding  problem  and  find  maximum  value  h  can 
have. 

13.  In  solving  16  12  —  800  1  +  h  =  0,  two  roots  are  obtained  ; 
find  the  sum  of  these  roots  and  the  product.     Interpret  the 
sum,  i.e.  give  the  physical  meaning. 

14.  Find  the  nature  of  the  roots,  without  completely  solving, 
in  the  following  equations  : 

a.   xi  +  3x-5  =  0.  c.   x2  +  3x  -  40=  0. 

5.   x*  _|_  3  x  -  8  =  0.  d   x*  -  3  x  +  40  =  0. 

e.   x2-3z  +  f  =  0. 

/.  4  x2  -12  x  +9  =  0. 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        95 

15.  Determine  the  nature  of  the  roots  : 

a.  4*2-  16*  -160  =  0.  c.   3  y2  +  16u  +  20  =  0. 

b.  7P  +  16«-  160=0.  d.   3v2  +  16v  +  25  =  0. 

16.  Plot  the  graphs  of  the  functions  in  15. 

17.  Solve  the  equations  of  15  graphically,  using  the  inter- 
section with  y  =  t2  or  y  =  y2  (one  half-inch  may  be  taken  for 
10  units  on  the  vertical  axis). 

18.  Find  the   sum   and   the   product   of   the  roots   in  the 
problems  of  14  and  15. 

f  x2  -\-  w2  =  36 

19.  Solve    {  '  by  substitution. 


g  __ 

20.  Solve    I         _      'by  substitution.     Draw  graphs. 

I  s  ==  ot  —  o, 

21.  Solve  .1  t2  —  50  1  —  30  =  0,  to  2  places  of  decimals. 

22.  Solve  t2  —  50  1  —  .0001  =  0  to  2  places  of  decimals. 

23.  Time  yourself  in  solving  the  following  10  quadratics, 
writing  the  roots  in  simplest  form  but  not  approximating  the 
square  root. 

a.  2  x2  +  3  x-  5  =  0.  /.   9  x2  =  20  x  +  10. 

b.  3z2-2z  +  7  =  0.  gr. 

c.  5;y2  +  12z  +  3  =  0.  h. 

d.  ?/2-3  y-  7  =  0.  t. 

e.  2v2 


24.   Time  yourself  in  finding  to  one  decimal  place  the  roots 
in  the  above  10  equations. 

8.    Equations  reducible  to  quadratics.  —  The  solution  of 

ax2  +  bx  +  c  =  0 

is  a  value  of  the  variable  x,  which  when  it  is  substituted  in 
cu»2  -\-bx-\-c,  makes  the  expression  0  ;  similarly  this  solution 
gives  a  value  of  the  variable  v,  or  t,  or  p,  or  £2,  or  f2,  or  3  tz  —  1, 
or  7  t2  +  2  t  —  3,  which  makes  the  expression  of  the  same  form 


96  UNIFIED  MATHEMATICS 

in  that  variable  zero ;  viz.,  a  value  which  makes  av*  -f  bv  4-  c 
equal  zero  when  the  value  is  put  for  v,  or  a(t2)2  +  bt2  -+-  c,  equal 
zero  when  the  value  is  put  for  t2,  or  a(3  tz  —  I)2  +  b(3  P—  1)  +  c 
equal  zero  when  the  value  is  put  for  3 12  —  1.  Equations 
which  can  be  put  in  the  form  cue2  +  bx  +  c  =  0  are  called 
equations  in  quadratic  form,  the  term  being  applied,  in  gen- 
eral, to  expressions  which  are  not  quadratics  directly  in  the 
principal  variable.  Thus,  in  any  expression  involving  x,  x2,  x3, 
x*,  xb,  or  x6,  the  value  of  the  expression  depends  primarily  upon 
the  principal  variable,  x ;  an  expression  like  3  x4  —  2  x2  —  7, 
involving  the  variable  x2,  its  square,  and  constants  as  coeffi- 
cients, is  said  to  be  in  quadratic  form,  and  it  is  a  quadratic  in 
the  variable  x2,  but  a  quartic  in  x. 

9.   Illustrative  exercises. 

1.    Solve  3  t*  -  5  t2  -  7  =  0. 

As  a  quadratic  in  t2,  the  formula  for  the  solution  of  a  quadratic  gives 


:  =  5±V109,  whence 


6  6 


There  are  four  values  represented  here,  of  which  two  are  imaginary. 

2.    Solve  x3  =  1,  or  x3  —  1  =  0,  and  x3  —  8  =  0  ;  these   illus- 
trate a  type  of  equation  reducible  to  a  quadratic  by  factoring. 

x»  -  1  =  (z  -  1)  (a?  +  x  +  1)  =  0. 

x  -  1  =  0,  x  -  1 


3  2 

These  values  —  —  —  —  ^—  ,   —  —  —  —  ^—  and  1  are  called  the   cube 
2  2 

roots  of  unity  ;  note  that  V—  3  is  denned  as  a  quantity  whose  square  is 
—  3  ;  the  systematic  discussion  of  such  numbers  is  deferred  until  a  later 
chapter.  Squaring  either  of  the  two  imaginary  cube  roots  of  unity  gives 
the  other  ;  these  roots  may  then  be  designated  as  1,  w,  w2.  The  cube 
roots  of  8  are  2,  2  w,  and  2  w2  •;  of  7  are  7*,  7^  w?,  and  7^  w>2,  wherein  7* 
denotes  the  real  cube  root  of  7. 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS        97 

PROBLEMS 

1.  Solve  for  t\  and  then  for  t.     t6  —  1  ?  —  8  =  0. 

2.  Solve  and  check  by  substitution  : 

or4  +  3  x~2  -  5  =  0. 

3.  2x*-7x*-5  =  0. 
4. 

Note  that  this  expression  when  cleared  of  fractions  gives 

x4  +  xs  +  x2  +  a;  +  1  =  0, 

a  factor  of  x5  —  1  =  0  ;  the  imaginary  roots  of  x  which  are  obtained  by 
solving  are  the  other  four  fifth-roots  of  unity. 

5.  (3  x2  _  S)2  +  2(3  x2  -  5)  -  7  =  0. 

6.  v  +  v*  =  10. 


7 
' 


8.  Find  the  value  of  x?,  and  of  x  in 

a?  +  3  x*  -  7  =  0. 

9.  Find  the  value  of  x2,  in 

x  _  3  ar*  -  7  =  0, 

and  compare  with  the  preceding.  The  real  test  of  a  value 
found  as  a  root  is  obtained  by  substituting  the  value  in  the 
given  expressions.  Squaring  may  introduce  a  new  root  ;  thus 
squaring  x  =  2,  gives  x2  =  4,  or  is  equivalent  to  multiplying 
x  —  2  =  0,  member  by  member  by  x  -f  2. 


10.   3  x  +  Vx  +  5  =  7.  11.   3  x  -  Vx  +  5  =  7. 

10.  Limiting  values  of  a,  b,  c.  —  As  c  approaches  more  and 
more  nearly  to  zero  as  compared  with,  a  and  b,  it  is  evident 
that  some  value  of  x  also  near  to  zero  will  satisfy  the  equation 
ax2  +  bx  +  c  =0  ;  this  value  will  be  of  the  same  sign  as  c  if  6  is 
negative,  and  opposite  in  sign  to  c  if  b  is  positive.  Thus 


98  UNIFIED   MATHEMATICS 

3x*  —  2x  —  .000001  =  0  is  an  equation  which  will  be  satisfied 
by  a  value  of  x  very  near  to  —  .0000005  ;  substituting  the  value, 
-.0000005  we  have  .00000000000075  +.000001  -  .000001  which 
is  surely  near  to  zero ;  even  if  a  were  very  large  compared 
with  b,  this  expression  has  one  root  near  to  zero.  Solving 

3  a;2 -2  a -.000001  =  0, 


_  2  ±  V4  +  .000012 
~6~ 

2  ±  2.000003     4.000003           .000003 
= —  =  —         —  or 

666 
=  |  or  -  .0000005. 

Similarly  in  1000  x2  —  3000  x  —  1  =  0,  one  solution  will  be  small, 
approximately  3-^0.  When  c  =  0,  the  roots  of  ace2  +  bx+  c  =  0 
are  the  roots  of  ax2  +  bx  =  0,  giving  a(acc  +  6)=0;  whence 

#=0  and  x  =  — .     When  both  b  and  c  approach  zero,  both 

a 
roots  of  the  quadratic  ax2  +  bx  +  c  =  0  approach  zero. 

When  a  approaches  zero  in  comparison  with  b  and  c,  one 
root  of  the  quadratic   becomes  very  large  and  the  other  ap- 
proaches — -.     Thus  in  the  quadratic 
b 

x>-  1000  x  -3000  =  0, 


x  =  1000  ±  VlOOOOOO  + 12000  =  1000  ±  1006  =  1003  or  _  3 

2  2 

/1 000  ±  1005.982 

— - —         -    are  the  more  exact  values,  giving  1002.991 
V  2 

or  -  2.991.") 

As  a  approaches  nearer  and  nearer  to  zero  one  root  becomes 
larger  and  larger  without  limit.     Thus  if  above  we  had 

.001  x2  -  1000  x  -3000  =  0 


_  1000  ±  VlOOOOOO +  12  _  1000  ±  1000.006 

.002  .002 

=  1000001.5  or  -  3  (more  exactly  2.991  as  before). 


THE  LINEAR  AND  QUADRATIC  FUNCTIONS      99 

Both  roots  become  large  if  both  a  and  b  become  small  as  com- 
pared with  c. 

PROBLEMS 

Find  first  approximate  values,  and  verify  by  solving  the 
quadratic  : 

1.  3x2-  7x  -.0001=0. 

2.  5xz  —  7x  —  .1  =0. 

3.  5  x2  —  .007  x  -  .001  =  0. 

4.  4000  =  3000  1  -  16  12  ;  one  root  is  the  number  of  seconds 
for  a  bullet  to  rise  4000  feet,  initial  velocity  3000  feet  per 
second,   air  resistance   neglected  ;  what   does  the  other  root 
represent  ? 

5.  .01  x*-  -  300  x  -  500  =  0. 

6.  .003  12  +  2  1  —  42  =  0  ;  this  gives  a  more  exact  equation  for 
the  temperature  at  which  the  velocity  of  sound  in  air  becomes 
40  feet  greater  than  it  is  at  0°  C. 

7.  1000000  x2  -  3000000  x  -  5  =  0. 

REVIEW  PROBLEMS 

1.  Plot  the  graph  of  y  =  3  x  —  5. 

2.  Plot  the  graph  of  the  following  functions  : 

a.  y  =  x2  —  4  x  +  5. 

b.  y  =  x2  —  •  4  x  +  4. 

c.  y  =  xz  —  4  x. 

d.  ?/  =  a;2  —  4  a;  —  2. 

3.  For  what  values  of  x  is  y  equal  to  0  in  the  four  functions 
of  the  preceding  question  ?     The  graphical  solution  is  desired. 

4.  Plot  15  points  from  x  =  —  .5  to  x=  +8  and  join  by  a 
smooth  curve  representing 


for  what  values  of  x  is  y  equal  to  zero  ? 


100 


UNIFIED  MATHEMATICS 


5.  s  =  20  t,+  50  —  16 12.     This  equation  represents  the  mo- 
tion of  a  body  thrown  from  a  height  of  50  feet  straight  up  into 
the  air  with  a  velocity  of  20  feet  per  second.     Plot  the  graph 
and  locate  the  position  of  the  body  at  the  end  of  1  second ;  at 
the  end  of  5  seconds. 

6.  Plot  the  graph  of  s  =  800 1  —  16 12,  for  values  of  t  from 
0  to  50  ;  note  that  it  is  desirable  to  get  the  values  of  s  first  for 
intermediate   values   and   to   choose   the  y-scale  accordingly. 
This  equation  represents  approximately  the  height  after  t  sec- 
onds of  a  bullet  shot  straight  into  the  air  with  a  velocity  of 

800  feet  per  second. 

7.   Plot     the      graph     of 


x-axis  and  7/-axis  off  the  paper 


-2t 


-140 


v= 


between  d  =  12  and 


Shifted  lines  of  reference 


d  =  20,  taking  the  scales  so 
as  to  enable  you  to  read  vol- 
umes as  correctly  as  possible 
within  these  limits.  Plot 
only  values  above  100  on  the 
y-scale,  and  to  the  right  of 
12  on  the  x-scale.  This  gives 
the  volume  in  cubic  units  per 
unit  of  height  of  cylindrical 
containers  which  have  radii 
varying  from  12  to  20  units. 
Apply  this  to  cans  and  to 
silos. 


8.  Plot  the  graph  of  t?  =  ^— ;  this  gives  the  time  of  beat  of 

a  pendulum  I  centimeters  long  where  gravity  is  980  cm.  per  sec. 
per  sec. 

9.  Plot  the  graph  of  y  =  as*,  for  values  of  x  from  0  to  8. 
10.   Plot  the  graphs  of  the  following  linear  functions  : 

a.  y  =  3  x  —  5.  c.   v  =  10  +  8 1.          e.    s  =  100  —  40 1. 

b.  y  =  3  x.  d.   s  =  6  -  3 1.          f.    y  =  —  2  x  + 10. 


.CHAPTER   VI 

STRAIGHT  LINE  AND   TWO-POINT  FORMULAS 

1.  Slope-intercept  formula  :  y  =  mx  +  k. 

The  equation  y  =  mx  +  k,  into  which  form  the  equation  of 
any  straight  line  can  be  put,  is  called  the  slope-intercept  form 
of  the  equation  of  a  line  ;  m  represents  the  slope  of  the  line  and 
k  is  the  intercept  on  the  y-axis.  The  equation  of  a  line  parallel 
to  the  ?/-axis,  x  =  k,  cannot  be  placed  precisely  in  this  form,  as 
the  ^-intercept  is  infinite. 

2.  Point-slope  formula  :  y  —  y\  =  m(x  —  x^. 

As  it  is  frequently  desired  to  find  the  equation  of  a  line  of 
given  slope  and  passing  through  a  given  point,  a  separate 
equation  in  terms  of  the  slope  and  coordinates  of  the  given 
point  is  desirable.  Let  the  equation  of  the  line  be  conceived 
as  in  the  form,  y  —  mx  +  k ;  since  (xl}  y^)  is  on  the  line, 
yl  =  mx!  -f-  k ;  subtracting  gives  y  —  yv  =  m(x  —  o^),  the  equa- 
tion of  the  straight  line  in  terms  of  m,  the  given  slope,  and 
(#!>  y\)  the  coordinates  of  the  given  point. 

3.  Two-point  formula:  ¥—£  =  y2  ~  Vl • 

X  —  X±         Xn  —  X\ 

The  equation  of  the  straight  line  through  (x1}  yi)(x2,  ?/2)  is 
also  easily  derived  from  the  slope-intercept  form. 

As  before  ?/!  =  mo^  +  k, 

yi  =  mx2  +  k, 
whence  y2  —  y1  =  m(x2  —  o^), 

and  m  =  ^2  ~  -^ ,  giving  m,  the  slope 

X2-Xi 

of  the  line,  in  terms  of  a/1}  y1}  x2,  and  y2. 

101 


102  UNIFIED  MATHEMATICS 

Hence,  y  —  yl  =  ^2  ~  ^1  (x  —  a^)  is  the  equation  of  the  line 
x%  —  Xi 

in  a  form  involving  only  the  given  constants. 

The  expression,  m  =  **  ~~  *l ,  represents  the  slope  of  a  line 

joining  (xi,  y^)  to  (xz,  yz).  Similarly  - — —  represents  the  slope 
of  the  line  joining  any  point  (a;,  y)  to  (xt,  yi).  The  preceding 

equation  of  the  line  in  the  form  - — —  =  & — —  is  an  equality 
of  two  slopes.  l 

The  formula  m  =  ^2  ~  •^1   is  frequently  used ;  it  should  be 

memorized  with  the  aid  of  the  diagram  as  placed  in  quadrant 
I.  This  formula  gives  the  rate  of  increase  of  y  in  the  interval 
from  (a^,  y±)  to  (a^,  t/2)  as  compared  with  the  increase  of  x  in 
the  same  interval ;  it  compares  the  change  in  y  in  the  interval 
with  the  change  in  x  in  the  same  interval. 

PROBLEMS 

1.  Find  the  equation  of  the  line  of  slope  3  and  y-intercept 
5 ;  with  m  =  3,  k  =  —  5  ;  m  =  —  3,  fc  =  8 ;  m  =  0,  fc  =  4 ;  ra  =  5, 

2.  Put  the  following  equations  into  slope-intercept  form : 

a.  3y  —  2 a;  +  5  =  0.  d.   y  —  3x  —  7  =  0. 

b.  3x  +  2y-7  =  Q.  e.    y  +  5  =  0. 

c.  x  +  2  y  =  0.  /.    x  +  3  =  0. 

3.  Write  the  equation  of  the  straight  line  through  ( —  2,  5) 
and  (1,  4)  ;   through  (3,  —  5)  and  (2,  1).     Find  intercepts  on 
both  axes  and  the  slope  in  each  case. 

4.  Write  the  equation  of  the  straight  line  through   (1,  5) 
having  the  slope  3.     Find  the  x  and  y  intercepts. 

5.  Find  the  equation  of  the  straight  line  through  (a,  0)  and 
(0,  6),  i.e.  the  line  having  intercepts  a  and  6,  respectively, 


STRAIGHT  LINE  AND  TWO-POINT  FORMULAS     103 

/j»          AI 

and  put  this  equation  into  the  form  -  +  "  =  1.     This  is  called 

a     b 
the  intercept  form  of  the  equation  of  a  straight  line. 

6.  Given  9  C  =  5  F  —  160,  the  formula  connecting  centi- 
grade and  Fahrenheit  readings  of  temperature,  find  the  slope 
and  the  x  and  y  intercepts.     Find  the  slope  of  the  line  join- 
ing (32,  0)  to  (212,  100).     What  is  the  rate  of  change  of  C  in 
the  interval  as  compared  with  the  change  in  F?     What  physi- 
cal meaning  have  the  intercepts  ? 

7.  Given  that  1000  cu.  cm.  of  mercury  at  0°  C.  increases  to 
1007.2  cu.  cm.  at  40°  C.,  find  the  rate  of  change  of  volume  per 
degree  of  temperature,  and  finally  per  cu.  cm.     Note  that  it  is 
not  necessarily  true  that  this  rate  found  for  an   interval   of 
40°  C.  should  be  the  uniform  rate  everywhere  in  the  interval. 
Write    the    equation    representing    the   volume   in   terms   of 
temperature,  assuming  that  the  relation  is  linear,  i.e.  that  the 
increase    in     volume    is    proportional    to    the    temperature. 
Mercury  expands  differently  at  different  temperatures,  but  the 
variation  is  slight  in  the  interval  from  0°  to  40°,  not  varying 
by  more  than  |  of  1  %  from  .00018  cu.  cm.  per  degree  for  1  cu.  cm. 

8.  Join  (0,  0)  to  (100,  39.37)  and  interpret  for  converting 
centimeters  to  inches  and  inches  to  centimeters ;  what  is  the 
meaning  of  the  slope?     Find  the  value  in  inches  of  18  cm., 
39  cm.,  47  cm.     Note  that  100  cm.  =  39.37  inches. 

9.  59.8  pints  of  water  weigh  approximately  62.4  Ib.     Draw 
the  graph  connecting  (0,  0)  to  (59.8,  62.4)  which  will  give  the 
approximate  weight  of  any  given  number  of  pints  of  water. 
How  could  you  read  the  weight  of  quarts  or  gallons?     Use 
1  inch  for  10  units  on  both  scales,  in  plotting. 

10.    Find  the  equations  of   the    straight  lines  joining  the 
following  pairs  of  points,  timing  yourself : 

a.  (3,  5)  to  (-  2,  7).  e.  (0,  8)  to  (0,  5). 

6.  (3,  5)  to  (2,  -7).  /.  (1,  -  3)  to  (-  1,  -  5). 

c.  (0,  8)  to  (7,  0).  g.  (1,  -  3)  to  (1,  6). 

d.  (0,  8)  to  (7,  -  6).  h.  (-1,  -3)  to  (-3,  -5). 


104 


UNIFIED  MATHEMATICS 


i.  (-1,  -3)  to  (3,  3). 

j.  (8,  -  3)  to  (- 3,  2). 

k.  (-3,5)  to  (7,0). 

I  (-3,  5)  to  (-7, -2). 


m.   (-  3,  5)  to  (7,  -  2). 
(2,  2)  to  (-2,  -2). 
o.   (100,  60)  to  (0,  0). 


n. 


11.   Find  the  equations  of  the  following  lines : 


a.  of  slope  3,  y-intercept  5. 

b.  of  slope  3,  ^-intercept  5. 

c.  of  slope  —  2,  y-intercept  0. 


d.  of  slope  —  1,  through  (2, 2). 

e.  of  slope  +  4,  through  (2, 2). 
/.    having  intercepts  of  7  and 

5  on  the  x-  and  r/-axes 
respectively. 


o  =  yz—  y\, 


4.    Distance  between  two  points  :   d  =  V(xz  —  j^)2  +  (y2  — 


Since 


Whatever  the  positions  of  Pl  and  P2,  parallels  to  the  x-  and 
y-axes  through  Pl  and  P2  form  a  rectangle  (a  straight  line  if 

*i/i  ==  3/*>  or  Vj  ^s  v->)  WIIOSG 
sides  are  in  absolute  value 
|  x2  —  xl  |  and  |  y2  —  yl  \  ; 
the  bars  indicate  that 
only  the  numerical  value 
is  considered.  As  a  posi- 
tive distance  P\M^  if 
Xi  >  x2,  would  have  to  be 
written  xv  —  x2 

But  since  the  numerical 
value  of  the  expression 
(x2  —  o^)2  is  the  same  as 


Distance  between  two  points 
d*  =  (zj  -  a*)'  +  (y*  -  2/i)2- 


the  value  of  (a^  —  a;2)2  we  may  use  in  every  case  (x2  —  x^)  for 
in  the  above  expression  for  d  wherein  only  the  square  of 
enters. 

d  = 


STRAIGHT   LINE  AND  TWO-POINT  FORMULAS     105 


The  distance  from  any  point  (xl}  yt)  to  any  point  (0%,  y2)  is 
given  by  this  formula  ;  this  distance  is  taken  in  general  as  a 
positive  quantity. 

d  This  formula  may  be  used  to  derive  the  equation  of  the 
straight  line  joining  PI(XI,  y\~)  to  P2(x.2,  y.2)  for  any  point  P(x,  y) 
on  the  line  is  such  that  PPV  +  P\Pz  —  PPz  5  and  for  no  point 
not  on  the  line  is  this  relation  true. 


5.    Point  of  division  formula  : 

_  *iX2  4-  Mi  .  .  .  _  *u 

5/2  +  fc^ 

4  ~~    til,     '  y'A  ~ 

K!'  +  Kn                                i 

fcl+*2 

A     A  A             A 

II                  >                         II 

P3  r*                       1*3  A 

1              > 

For  any  three  points  P,,  P2,  and  P3  on  a  directed  line  we 
have  PXP2  4-  P2P3  =  PiP9  5   if  P  lies  between  Pl  and  P3,  all 

three  segments  have  the  same  algebraic  sign  but  otherwise 
positive  and  negative  segments  are  involved. 

OPi  +  PiP-2  =  OP2  is  then,  similarly,  the  fundamental  rela- 
tion true  for  any  three  points  on  a  directed  line,  whence 

PXP2  =  OP2  -  OP!  =  x2  -  *!. 

In  words  the  distance  on  the  avaxis  (or  any  other  line  parallel 
to  the  as-axis)  from  any  point  whose  abscissa  is  a^  to  any  point 
whose  abscissa  is  x2,  is  given  by  xz  —  xt.  Similarly  with 
respect  to  points  on  the 
y-axis,  or  two  points  on  a 
line  parallel  to  the  t/-axis, 
the  distance  from  the 
point  whose  ordinate  is  yl 
to  the  point  whose  ordi- 
nate is  y%  is  ?/2  —  y^. 

To  find  the  coordinates 
of  the  point  P3  which 
divides  the  line  joining 
P]P2  into  two  segments 
which  bear  to  each  other 

T. 

the  ratio    -*,    note    that 


-1! 


-"A 


Point  of  division  formula 


drawing  lines  through  P,,  P2,  and  P3  parallel 


106 


P  /-*       k 

1  if  3_£H. 

-43X2        A'o 

to  the  axes,  similar  triangles  are  formed,  or  the  proposition  of 
plane  geometry  that  a  series  of  parallels  cut  off  on  trans- 
versals proportional  parts  may  be  directly  used. 


Whence 


oc3  —  a?,  _ 


A3A2 
fc, 


Similarly,  3  =     ;  whence 

' 


2/2-2/3 


,    -, 
H  — 


Wherein  r  =   l. 


If  P3(a^,  2/3)   divides  the  line  PiP2  externally  in  the  ratio 

Jc 

—  ,  or  r,  the  segments  must  be  regarded  as  of  opposite  signs  and 

k*  j. 

consequently,  the  ratio  —  J  ,  or  r,  is  negative.     Either  k\  or  A*2  can 

«2 

be  regarded  as  negative  ;  shifting  the  sign  from  A"2  to  ^  is 
equivalent  to  changing  the  sign  of  the  numerator  and  denom- 
inator in  the  value  of  x%  and  2/3,  no  change  is  necessary  in  our 
above  derivation  of  the  values  of  a^  and  y3. 

By  eliminating  kt  and  Jc2  between  the  two  equations, 


+  k 


the  equation 
tained. 


of  the  straight   line  joining  P1  and  P2  is  ob- 


STRAIGHT LINE  AND  TWO-POINT  FORMULAS     107 

Mid-Point  : 

Place  k{  =  Jc2,  or  place  r  =  1, 


_ 


This  mid-point  formula  is  of  such  frequent  use  that  it 
should  be  separately  memorized  ;  the  truth  of%  it  is  obvious 
from  the  figure. 

PROBLEMS 
1.   Plot  the  locus  of  each  of  the  following  equations  : 


x  -  y  -  8  =  0. 

Plot  the  two  graphs  on  one  diagram  with  reference  to  the 
same  system  of  axes.  Locate  the  point  of  intersection,  graph- 
ically and  analytically. 

2.  Plot  the  graph  ofy  =  x3  —  3xi  —  8x  —  2.     Discuss. 

3.  Plot  the  graph  of5y+2z-5  =  0. 

4.  Plot  p  =  51  +  50. 

5.  Find  the  equation  of  the  straight  line  joining  A(—  2,  5) 
to   B(3,  7).     Find  slope   of   this   line.     Find   length   of  AB. 
Find  the  point  of  trisection  nearest  A.     Find  a  point  on  BA 
extended  that  divides  the  segment  BA  externally  in  the  ratio 
1:2. 

6.  Given  that  the  velocity  of  sound  at  0°  C.  is  1090  feet 
per   second,  and  at  30°  C.  is  1150  feet  per  second,  find  the 
velocity  at  20°  C.,  assuming  that  the  relation  is  linear  ;  the 
point  dividing   the   line  joining  (0,  1090)  and  (30,  1150)  in 
the  ratio  2  :  1  will  give  the  velocity  as  the  ordinate.     At  what 
temperature  will  the  velocity  be  1100  ft.  per  second  ?     What 
are  the  velocity  and  temperature  at  the  middle  point  of  the 
range  given? 


108  UNIFIED  MATHEMATICS 

7.  The   resistance  of   wire   increases   uniformly  with  the 
temperature,   r  =  r0(l  +  at),   the   rate   of    increase   depending 
upon  the  material  of  the  wire  ;  r0  is  the  resistance  at  0°  C.  and 
a  is  a  constant.     If  a  given  piece  of  wire  has  a  resistance  of 
200  ohms  at  10°  C.  and  of  208.4  at  30°  C.,  find  the  resistance  at 
the  middle  point  [of  (10,  200)  and  (30,  208.4)].     Find  the  equa- 
tion for  r  in  terms  of  L     Find  the  value  of  r  when  t  =  0  ; 
interpret  ;  find  the  value  of  t  when  r  =  0.     The  theory  is  that 
at  a  temperature  of  absolute  zero  (—  273°  C.  or  thereabouts) 
the  resistance  would  be  zero.     Ans.   r  =  195.8  -|-  .42  t. 

8.  The  resistance  of  copper  wire  of  fixed  diameter  varies 
with  the  length.     If  the  resistance  9f   1450  feet  of  a  given 
spool  is  184  ohms,  and  the  resistance  of  0  feet  is  0  ohms,  find 
the  equation  for  r  in  terms  of  1.     Plot  (0,  0)  and  (1450,  184). 
What  would  be  the  resistance  of  5280  feet  of  this  wire  ? 

9.  Between  (—1,  5)  and  (8,  37)  insert  9  points  dividing 
the  line  into  ten  equal  parts,  using  the  formulas 

x  =  kte  +  frga?!  and         J<Wz±]Wi 

KI  -}-  K%  KI  -J-  #2 

rearranged  as  follows  : 


x  _    -i     -    ii        i2  ~    ii  _  x    I         1     /x  _  x  \ 

KI  -f-  Ar2  fci  -f-  A*2 

and  similarly  fci       , 

y  =  yi  +  i  —  rirdfr-yi)- 

KI  -j-  K% 

Note  that  fcx  +  k2  is  constant,  10,  and  fcx  changes  for  the  nine 
points  from  1  to  9.  Use  this  method  in  problems  10,  11,  and 
14-17  below. 

10.  Between  (21,  .3584)  and  (22,  .3746)  insert  5  values  di- 
viding the  interval  into  6  equal  parts. 

11.  Between  (10,  .3611)  and  (20,  .3638)  insert  9  values  di- 
viding the  line  joining  these  points  into  ten  equal  parts. 

12.  Find  the  point  P3  dividing  the  line  joining  Pj(—  1,  5) 
to  Pz(8,  37)  externally  in  the  ratio  1  to  7  ;  externally  in  the 


STRAIGHT  LINE  AND  TWO-POINT  FORMULAS     109 

ratio  yL ;  externally  in  the  ratio  ^.     Note  that  either  ki  or  k2 
must  be  made  negative,  or  r  taken  as  negative. 

13.  Find  the  point  dividing   (21,  .3584)  to  (22,  .3746)  ex- 
ternally in  the  ratio  ^-,  f ,  |. 

14.  logl  =  0;  log 2  =.301;  find   9  values   dividing   (1,   0) 
and  (2,  .301)  into  ten  equal  parts.     Compare  with  the  loga- 
rithms of  1.1,  1.2,  1.3,  1.4,  •••  1.9.     See  problem  9  above. 

15.  log  200  =  2.3010  ;  log  210  =2.3222;  insert  9  values  be- 
tween (200,   2.3010)   and    (210,  2.3222),  comparing  with   the 
logarithms  of  201,  202,  •••  209. 

16.  log  200  =  2.3010  and  log  201  =  2.3032 ;   insert  9  values 
between  (200,  2.3010)  and  (201,  2.3032)  and  interpret. 

17.  Given  y  —  32  x  —  17  ;  find  the  corresponding  values  of 
y  when   x  =  10   and  x  =  20.     Find  the   points  dividing   this 
line  in  the  ratio  ^,  ^,  ^,  -|,  T.     What  points  of  division  are 
obtained  ? 

18.  Given   PI(—  1,  5)   and   Po(8,  37) ;  on  the   line  joining 
these  two  points  find  the  point  whose  abscissa  is  3,  without 
finding  the  equation  of  the  line.     Find  the  point  whose  ab- 
scissa is  7.     Find  the   point  whose  ordinate  is  16.     Find  the 
point  whose   ordinate   is   0.     Find  the  point  whose   abscissa 
is  +14. 

_    1        1       O 

19.  Eliminate  r  between  the   two  equations   x  = — 

1  +  r 

and   y  =  —        —•    These  two  equations  constitute  what  are 

known  as  parametric  forms  of  the   equation  of  the  straight 
line  joining  (—  1,  5)  to  (8,  37). 

20.  Write   the   equations   of  the  line   joining   (3,  —  2)  to 
(15,  28)  in  parametric  form.     Find  6  points  on  this  line. 


CHAPTER   VII 


TRIGONOMETRIC   FUNCTIONS 

1.  Angles  and  angular  measurement.  —  The  angle  made  by 
any  line  OP  with  the  horizontal  line  OX  is  regarded  as 
generated  by  a  moving  line,  an  arm  or  ray,  starting  from  the 
position  OX  and  turning  about  the  point  0  as  a  pivot,  moving 

always  in  the  same  plane.  This 
moving  ray  if  rotated"  in  the  sense 
contrary  to  that  in  which  the 
hands  of  a  clock  move,  counter- 
clockwise, is  regarded  as  gener- 
ating a  positive  angle  ;  clockwise 
rotation  generates  a  negative 
angle.  A  natural  unit  of  angular 
magnitude  is  the  complete  rota- 
tion which  brings  the  moving  arm 
back  to  its  original  position.  This 
Angle  generated  by  rotation  measure  is  used  in  giving  the 

speed  of  rotation,  e.g.  the  angular 

speed  of  rotating  shafts  and  wheels  is  measured  in  revolutions 
per  minute  or  per  second.  The  angle  generated  when  the 
moving  ray  is  in  the  same  straight  line  with  its  original  posi- 
tion, but  extending  in  the  opposite  direction,  is  called  a  straight 
angle  ;  half  of  this  angle  is  the  right  angle,  which  was  probably 
the  earliest  measure  of  angles  used.  Thus  our  terms  acute  and 
obtuse  relate  to  the  right  angle.  If  the  angle  is  conceived  as 
given  by  the  relative  position  of  two  lines  non-directed,  it  is 
evident  that  only  angles  less  than  a  straight  angle  would  be 
discussed. 

110 


TRIGONOMETRIC  FUNCTIONS  111 

In  the  ancient  development  of  geometry  the  right  angle,  so 
necessary  in  building,  was  fundamental ;  in  Greece  up  to 
about  150  B.C.  the  right  angle  was  used  as  the  unit  of  measure. 
The  artificial  division  of  the  complete  rotation  into  360  equal 
angular  units  called  degrees  is  due  to  the  Babylonians,  who 
made  this  subdivision  as  early  as  1000  B.C.  The  Babylonians 
used  60  as  a  unit  of  higher  order  much  as  we  use  ten,  and  it  is 
probable  that  they  divided  another  natural  angular  unit,  one 
sixth  of  a  perigon,  given  by  the  easy  construction  of  a  regular 
hexagon,  into  60  equal  parts  called  degrees ;  each  degree  was 
divided  by  them  into  60  minutes  (paries  minutiae  primae,  in 
Latin,  whence  "  minutes ")  and  the  minute  into  60  seconds 
(partes  minutiae  secundae). 

Another  natural  system  of  measuring  angles  is  of  funda- 
mental importance  in  mathematical  work.  This  is  the  circular 
system,  in  which  the  unit  angle,  called  a  radian,  is  the  angle 
measured  at  the  center  of  a  circle  by  an  arc  whose  length 
is  the  radius.  The  radius  can  be  laid  off  on  the  circle 
2ir,  6|  or  6.2832,  times,  and  since  equal  angles  at  the  center 
are  intercepted  by  equal  arcs  on  the  circumference,  this  angle 
can  be  placed  2  -n-  times  around  the  center,  or  approximately 
6^  times  in  a  complete  revolution.  Just  as  1°  is  used  for  1 
degree,  so  lr  is  used  for  1  radian,  and  similarly  for  other 
numerical  values;  when  no  angle  sign  is  used  radians  are 
understood. 

2ir"  =  360°.  ^  =  60°. 

o 

7rr  =  180°.  -=30°. 

6 

^=90.  -  =  45°. 

2  4 

The  student  should  accustom  himself  to  expressing  angles 
in  radians,  particularly  the  angles  of  30°,  45°,  60°,  and  90C,  and 
those  which  depend  directly  upon  them. 

Thus  160°  =  —  ;  135°  =  5^ ;  or,  with  many  writers,  simply  —  desig- 

64  4 

nates  135°  in  radians. 


112 


UNIFIED  MATHEMATICS 


A  natural  system  of  measurement  of  angular  magnitude 
The  arc  AR  equals  the  radius  OA  •  arc  A'R'  =  OA';   arc  A" R"  =  OA". 


Radian  system  of  measuring  angles 

In  these  circles  it  is  true  that  - —  =  — — -=  . 

OB       OB        OB 


It  is 


evident  that  if  the  circumference  of  one  of  these  circles  is 
divided  into  any  number  of  equal  parts,  then  lines  from  0  to 
these  points  of  division  will  divide,  when  extended,  the  other 
circles  into  the  same  number  of  equal  parts.  The  angle  at 
the  center  is  measured,  we  may  say,  by  the  intercepted  arc ; 

fjT 

or  0  =  — ,  wherein  a  stands  for  the  length  of  the  arc  and  r  for 

r 


TRIGONOMETRIC  FUNCTIONS  113 

the  radius.  The  length  of  the  arc  is  given  by  the  formula, 
a  =  rO,  when  0  is  measured  in  radians ;  the  area  A  of  the  cor- 
responding sector  of  angle  0  is  A=\  rW.  Since  3.141593" =180°, 

1ftO° 

1"  =  — — —  =  57.29578°,  1°  =  .0174533  radian. 

2.    Quadrants.  — 

The  two  lines  OX  and  0  Y  divide  the  plane  into  four  quad- 
rants, numbered  as  indicated  on  the  preceding  diagram,  I  to 
IV;  we  will  commonly  designate  a  quadrant  by  its  numeral. 
In  trigonometric  work  we  conceive  angles  as  placed  with  the 
vertex  at  0,  one  arm  falling  upon  the  OX  axis  to  the  right,  and 
the  other  arm  falling  in  one  of  the  four  quadrants,  or  upon  one 
of  the  axes.  We  think  of  the  angle,  in  effect,  as  generated  by 
an  arm  rotating  about  0  from  the  initial  position  OX.  Under 
this  assumption  it  is  evident  that  the  terminal  arm  of  an  angle 
may  fall  in  any  quadrant  either  by  a  positive  rotation  or  by  a 
corresponding  negative  rotation,  the  difference  between  the 
two  angles  being  360°.  Rotations  of  greater  than  one  revolu- 
tion reproduce  in  order  the  positions  on  the  diagram  produced 
by  rotations  of  less  than  one  revolution,  e.g.  angles  of  30°, 
-330°,  +390°,+  750°,  -  690°,  and  in  general  terms,  n  x  360° 
+  30°  where  n  is  any  integer,  are  represented  by  the  same 
figure.  In  radians  we  may  say  that  ar  and  (2  mr  +  «r)  are 
represented  by  the  same  diagram  for  all  integral  values  of  n. 

PROBLEMS 

1.  Using  3|  for  TT,  compute  the   value   of   lr  in  degrees. 
What  is  the  percentage  error  ? 

2.  Using  3^  for  TT,  compute   the  value  of  1°  in  radians. 
Percentage  error  ? 

r        K  _r 

3.  Give  the  value  in  degrees  of  4-  revolution  ;  irr ;  — ;  - — ; 

6       6 

1  straight  angle  ;  |  of  one  right  angle  ;  3r ;  -  irr ;  — ;  ^->  "  r~] 

2344 


114  UNIFIED  MATHEMATICS 

4.  Give  the  value  in  radians  of  1  revolution  ;  180° ;  45° ; 
135° ;  60° ;  120°  ;  225° ;  3  right  angles  ;  390° ;  765°. 

5.  What  is  the  percentage  error  in  using  57.3°  as  the  value 
of  1  radian  ? 

6.  What  error  in  seconds  is  introduced  by  using  57.3°  for 
1  radian  in  finding  the  value  of  3  radians  ?    3r  =  171.9° ;  an 
error  of  1  °/0  would  be  approximately  1.7°. 

7.  A  bicycle  rider  pedals  at  the  rate  of  20  miles  per  hour ; 
how  many  revolutions  does  the  rear  wheel,  diameter  28  inches, 
make  per  minute  ?    The  rear  sprocket  wheel,  diameter  4  inches, 
makes  the  same  number  of  revolutions  as  the  rear  wheel ;  how 
many  revolutions   does   the  front   sprocket  wheel,  diameter 
10  inches,  make  ?     Changing  gear  shifts  the  chain  to  a  smaller 
rear  sprocket ;  what  speed  will  be  attained  at  the  same  rate  of 
pedaling  by  shifting  to  a  3-inch  rear  sprocket  ? 

8.  Place  the  following  angles  in  their  proper  quadrants : 

150°,  240°,  760°,  -  840°,    ^-,   — ,    -  ~  if.     Give  the  cor- 

o          4  «3 

responding  positive  angles  less  than  2  TTT. 

9.  In  the  circle  of  radius  10  what  is  the  length  of  the  arc 
of  an  angle  at  the  center  of  60°?     What  is  the  difference 
between  an  arc  of  60°  and  an  angle  of  60°?     WThat  is  the 

length  of  the  arc  of  30°,  45°,  -,  —  ? 

6      6 

10.  What  is  the  angle  at  the  center  in  radians  and  degrees, 
in  a  circle  of  radius  100,  subtended  by  an  arc  of  length  100  ? 
50  ?   30  ?    100  TT  ?     Find  the  areas  of  the  corresponding  sectors 
of  the  circle. 

11.  In  the  artillery  service  angles  are  measured  in  "mils"; 
a  "  mil "  is  defined  as   6^d  of  a  complete  revolution.     Com- 
pute the  value  in  radians  of  one  mil. 

12.  On  the  "  mariner's  compass  "  the  complete  revolution  is 
divided  into  32  parts,  called  "  points  "  of  the  compass  ;  com- 
pare the  "  points,"  with  degrees,  "  mils/'  and  radians. 


TRIGONOMETRIC  FUNCTIONS  115 

13.  Compute  the  value  of  the  "  mil "  in  minutes  and  give 
approximate   formulas   for  converting   "  mils "   into  minutes 
and  conversely. 

14.  At  what  rate  per  second  in  degrees,  and  in  radians,  do 
the  hands  of  a  clock  turn  ? 

15.  A  grindstone  of  diameter  18  inches  is  turning  246  times 
per  minute.     Compute  the  linear  velocity  of  a  point  on  the 
rim. 

16.  In   grinding  certain   tools   the   linear  velocity  of   the 
grinding   surface   should   not   exceed  6000   feet   per  second. 
Find  the  maximum  number  of  revolutions  per  second  of  a 
10-inch  (diameter)  emery  wheel  and  of  a  5-inch  wheel. 

17.  Find  the  angular  velocity  in  revolutions  and  in  radians 
of  an  Ohio  grindstone,  2  feet  in  diameter,  which  should  have  a 
circumferential  speed  of  2500  feet  per  minute. 

18.  The  path  of  the  earth  is  approximately  a  circle  with 
radius  93,000,000  miles ;  find  the  distance  traveled  in  1  day. 
What  percentage  of  error  would  be  introduced  by  using  365 
instead  of  3,65i  days  ?     Show  that  the  fact  that  we  give  r  as 
93,000,000  implies  that  the  position  of  the  point  on  the  earth 
would  not  affect  our  computation. 

3.  Polar  coordinates  and  angular  variables.  —  Any  point  P 
in  the  plane  may  be  located  by  giving  its  distance  from  a 
fixed  point  0,  called  the  pole,  and  the  angle  which  a  line  from 
the  pole  to  the  point  P  makes  with  a  fixed  line  OR,  called  the 
polar  axis.  In  general  terms  the  polar  coordinates  of  any 
point,  of  a  variable  point,  are  designated  by  r  and  6,  radius 
vector  and  vectorial  angle.  (See  p.  116.) 

r  will  be  assumed  to  be  a  positive  quantity,  and  0  may 
be  assumed  as  the  angle  generated  by  the  rotation  of  the 
vector  OP  from  an  initial  position  on  OR.  A  negative  angle 
is  generated  with  the  polar  axis  by  a  line  which  turns  from  the 
polar  axis,  about  0,  in  the  clockwise  direction.  Thus  the 
/-  R  OP  is  taken  as  +30°;  this  same  figure  may  also  be  con- 


116 


UNIFIED  MATHEMATICS 


Polar  coordinate  paper 
Location  f  (1°'  3°0)'     (10'   150&)'    (10'    -150°)'   and     (10'    - 

<"    U10-  f)- 


ceived  as  representing  —  330°.  Angles  which  differ  by  multi- 
ples of  360°,  generated  by  lines  rotating  from  an  initial  posi- 
tion upon  the  polar  axis,  are  represented  by  the  same  diagram  ; 
two  such  angles  are  commonly  called  "  congruent  "  angles. 
Each  rotation  of  360°  brings  a  line  back  to  its  starting  place. 

PROBLEMS 

1.   Locate  the  points  (3,  30°),  (6,  90°),  (4,  45°),  (8,  135°), 
(3,  270°),  (6,  -  90°),  (5,  180°),  and  (2,  390°). 


^  Locate  the  points  (&,  £\   /4,    £\   (6,  0),  (3,   -£\ 
),  and  (3,  3^).       V  V 


3.  What  is  common  to  all  points  on  OR  ? 

4.  What  curve  is  represented  by  r  =  10  ? 


5.   What  curve  is  represented  by  6  =  30°  or  6  =  —  ? 

6 


TRIGONOMETRIC  FUNCTIONS 


117 


"/r 


4.  Trigonometric  functions  —  sine  and  cosine.  —  Assume  an 
x-axis  to  coincide  with  the  polar  axis,  and  a  y-axis  to  be  drawn 
perpendicular  to 
the  polar  axis  at 
the  pole.  When 
0  is  any  fixed 
angle,  the  coordi- 
nates (x,  y)  in 
rectangular  coor- 
dinates and  (r,  ff) 
in  polar  coordi- 
nates, of  points 
upon  the  ray 
making  the  angle 
6  with  OX,  are 
connected  by  the 
following  relations  : 

—  =  —  =  —  =  - ,  for  points  upon  the  ray, 
TI      r2      rs     r 


-Va 


and  ^=^  =  ^  =  2?. 

1  2  3 

x2  +  y2  =  r2,   for  any  point  (x,  y)  in  the  plane. 

We  may  say  that  -   is  a  constant  for  any  given  angle  0; 
r 

this  constant  changes  as  0  changes.  It  is  evidently  a  function 
of  0.  Since  r  remains  positive,  this  function  is  positive  for 
all  angles  0  represented  in  the  upper  quadrants ;  negative  for 
angles  in  quadrants  III  and  IV.  This  constant  is  ^  f  or  0  =  30° 

or  .707  for  0  =  45°,  \ V3  or  .866  for  6  =  60°,  1  for 


or 


f)  2 


0  =  90°,  .866  for  6  =  120°,  .707  for  0  =  135°,  \  for  0  =  150°,  and  0 
for  6  =  180°,  all  by  elementary  geometry.  When  0  is  an  angle 
which  lies  in  quadrant  III  or  IV,  i.e.  values  of  0  between  + 180° 


118 


UNIFIED  MATHEMATICS 


and  -f  360°,  this  function  of  0  becomes  negative, 
of  0  is  called  the  sine  of  6,  or  sin  0. 


This  function 


1 


sin  9  =  "  • 

T 

cos*  =  -- 

r 


Polar  coordinates,  r  and  0 
Rectangular  coordinates,  x  and 


COS  6  = 


Similarly  the  ratio  - 
»" 

is  a  constant,  whose 
value  depends  entirely 
upon  the  position  of  the- 
moving  ray  ;  this  func- 
tion of  0  we  define  as 
cosine  0. 


sin0  = 


The  consideration  of  the  changes  in  value  of  these  functions  of 
0,  sin  0,  and  cos  0,  as  6  changes,  is  facilitated  by  thinking  of  the 
moving  ray  as  fixed  in  length. 

For  positive  values  of  0  less  than  90°,  6  in  I  or,  in  symbolic 
language,  0  <  0  <  90°,  it  is  evident  that  the  complementary 
angle  to  any  angle  0  gives  a  tri- 
angle similar  to  the  triangle  in- 
volving 0.  In  this  second  tri- 
angle the  ordinate  and  abscissa 
correspond  respectively  to  the  ab- 
scissa and  ordinate  in  the  original 


triangle,    whence    ?L  —  _  • 
r      r 


Now 


y~=  sin  (90°  -0),  and    ?=cos0; 

r  r 

hence  cos  0  =  sin  (90°  —  0),  or,  in 

words,  the  cosine  of  any  angle 

6  (0  <  0  <  90°)  is  the  sine  of  the 

complement  of  0.     This  explains 

the  name,  cosine  0,  which  is  simply  the  "  complement's  sine." 


Complementary  angles, 
9  +  9'  =  90° 


Further,  —  =  -  ,  whence  cos  (90°  —  0)=  sin  0. 
r      r 


TRIGONOMETRIC  FUNCTIONS  119 

Either  one  of  the  triangles  may  be  regarded  as  the  origi- 
nal, the  complementary  angle  will  be  found  in  the  other; 
the  demonstration,  as  given,  applies  in  either  case.  The 
above  figure  serves,  then,  to  demonstrate  the  two  formulas, 
sin  (90°  —  0)  =  cos  6  and  cos  (90°  —  0)  =  sin  0,  for  any  positive 
acute  angle  0.  Later  these  formulas  will  be  shown  to  hold  for 
all  angles  6,  without  restriction  as  to  magnitude  or  sign. 

The  formula  cos  (90°  —  $)  =  sin  0  may  be  derived  from 
'  sin  (90°  -  0)  =  cos  0  by  substituting  for  0  the  value  90°  -  0',  and 
finally  replacing  0'  by  0.  Since  0  may  vary  from  0  to  90°, 
90°  —  6  varies  between  the  same  limits. 

sin  (90°  -6)=  cose, 

cos  (90°  —  8)  =  sin  6. 

5.  Historical   note.  —  The   function   sin  6   is    Hindu   in   its 
origin,  dating  back  probably  to  the  fourth  century  A.D.     The 
Hindus  called  the  sine  "  ardha-jiva,"  meaning  half-chord.     In 
the  eighth  century  A.D.  the  Arabs  becoming  familiar  with  Hindu 
astronomy  and  trigonometry,  as  used  in  astronomical  work, 
transliterated  the  word  "jiva"  or  "jiba"  into  "geib";  the 
word  in  Arabic  means  curve  and  in  the  twelfth  century  Euro- 
pean translators  into  Latin  of  Arabic  works  of  science  trans- 
lated this  word  as  "  sinus."     Into  English  the  word  comes  by 
transliteration  again,  the  sound  and  not  the  sense  being  pre- 
served. 

Plane  trigonometry  is  possible  using  the  chords  instead  of 
the  half-chords  ;  this  system  was  developed  by  the  Greeks,  but 
it  leads  to  much  more  complicated  formulas  and  methods. 

6.  Tangent  and  the  reciprocal  functions.  —  The  quotient  —  — 

cos  0 

varies  as  0  varies  ;  this  is  then  a  function  of  0.     This  function 
is  called  the  tangent.   •  By  definition, 


tan  8  = 


cos 


Y 
xj 


120  UNIFIED  MATHEMATICS 

The  reciprocals  of  sin  6,  cos  0,  and  tan  6  are  also  functions  of 
0 ;  to  these  the  names  cosecant  0,  secant  0,  and  cotangent  0 
have  been  given. 

By  definition,    cogecant  ^  ^  e  =  _L_ 

sin6 

4 

secant  0,  or   sec  8  = 
cotangent  6,  or   cot  6  = 


cos  6' 

1 
tan  6 


PROBLEMS 

1.  Given     sin  0  =  .29,     find      cos  0    using     the     formula 
sin2  0  +  cos20  =  1.     The  negative  value  has  a  meaning. 

2.  In  what  quadrants  is  sin 9  positive?  in  what  quadrants 
is  cos  6  positive  ? 

3.  Given  sin  0  =  .29,  in  what  quadrants  may  6  lie  ? 

4.  In  what  quadrants  is  tan  0  positive  ? 

5.  As  a  rotating  arm  of  length  10,  moving  about  0  from 
OX,  turns  through  90°,  discuss  the  changes  in  value  of  the  y  of 
the  end  of  the  moving  arm ;  consider  r  as  10  and  discuss  the 
change  in  value  of  sin  0  as  the  angle  generated  increases  from 
0°  to  90°  to  180°.     What  change  in  sin  0  as  0  increases  beyond 
180°? 

6.  Discuss  similarly  the  changes  in  values    of  cos  6  as  6 
varies  from  0°  to  90° ;  from  90°  to  180°. 

7.  Discuss  the  possible  values  of  tan  0.     Take  x  =  1,  |-,  .1, 
.01,  .001  and  compute  y  in  a  circle  of  radius  10.     Discuss  the 
values    of   tan  0.     When    x  =  .000001,    y  =  9.99999999999995 
what  is  the  approximate  value  of  tan  6  ? 

8.  Given  tan  0=3,  find  sec  6  from  the  formula 

sec2  0  =  1  +  tan2  0. 
Compute  both  the  positive  and  the  negative  values  of  cos  6. 

9.  Express   in   terms   of   the   sine   of   the    complementary 
angle  :  cos  48°,  cos  84°,  cos  56°,  cos  48°  10',  cos  90°. 


TRIGONOMETRIC  FUNCTIONS  121 

10.  Express  in  terms  of  the  cosine  of  the  complementary 
angle  sin  48°,  sin  84°,  sin  56°,  sin  48°  10',  sin  90°. 

11.  Complete  the  following  table : 

cos  45°  =  .7071  =  sin  45° 
cos  46°  =  .6947  =  sin  44° 
cos  47°  =  .6820  =  sin 
cos  48°  =  .6691  =  sin 
cos  49°  =  .6561  =  sin 
cos  50°  =  .6428  =  sin 
Reverse  the  table,  beginning  sin  40°  = 

sin  41°  = 

12.  Complete  the  following  table  : 

sin  35°  =  .5736  =  cos 
sin  35°  10'  =  .5760  =  cos 
sin  35°  20'  =  .5783  =  cos 
sin  35°  30'  =  .5807  =  cos 
sin  35°  40'  =  .5831  =  cos 
sin  35°  50'  =  .5854  =  cos 
sin  36°  =  .5878  =  cos 

Notice  that  the  sines  of  35°  +  some  minutes  are  cosines  of 
angles  54°  +  some  minutes ;  the  cosines  of  35°  +  minutes  are 
sines  of  the  complements,  54°  +  minutes.  In  our  tables  you 
have  written  at  the  left  of  the  table  35°  and  54°  at  the  right ; 
sin  at  the  top  and  cos  at  the  bottom. 

35° 


1 

sin 

cos 

0 

.5736 

.8192 

60 

10 

.5760 

.8175 

50 

20 

.5783 

.8158 

40 

30 

.5807 

.8141 

30 

40 

.5831 

.8124 

20 

50 

.5854 

.8107 

10 

60 

.5878 

.8090 

0 

cos  sin 

54° 


122 


UNIFIED   MATHEMATICS 


7.    Fundamental  formulas.  —  Since  x*  +  yz  =  r2,  for  any  point 
on  this  circle  of  radius  r, 


1 


Note  that  although  x  or  ?/  or  both  may  be  negative,  the  rela- 
tion continues  to  hold  since  (—  x)z=.xz,  and  (—  y)z  =  y2. 
Whence,  by  substitution,  cos2  6  +  sin2  6  =  1,  for  all  values  of  6. 

By  division  by  cos2  0, 

sin2  6         1 


1  + 


cos2  0     cos2  0' 


or  1  +  tan2  0  =  sec2  0,  for  all  values  of  0. 
Similarly,  1  +  cot2  0  =  esc2  0. 

sin2  6  +  cos26  =  1. 
1  +  tan2  6  =  sec2  0. 


TRIGONOMETRIC  FUNCTIONS  123 

These    formulas    are   of    fundamental   importance.      They 
should  be  memorized. 


8.   Functions   oi 
plane  geometry  t 
determined   for   t 
structed   with  ru 
most   important 
45%  60°,  and  72° 
for  0  and  90°  (as  ] 

sin  45°  = 

cos  45°  = 
tan  45°  = 

0°,  30°,  45°,  60°,  and  r 
le  values  of  these  functic 
he  angles  which  can  be 
ler  and   compass.      The 
of  these  angles  are   30°, 
;   the  values  are  evident 
imits). 
1        v  2      7ft7 

el£ 

m 

8 

ited  angles.  —  By 
i  can  be  precisely 
eometrically  con- 

/ 

/ 

s 

/ 

z 

/ 

A/2       2 

1           A/2 

~^r~v=:i  7' 

I  =  cot  45°. 

/ 

i 

Functions  of  45° 

One  half  a  unit 
square. 

qin  fif>°           * 

. 

S 

/ 

—  cos  30° 

2f 

S 

*N 

^ 

1 

4 

._  /JAO            1 

s 

' 

/ 

JO 

\ 

cos  oO  =  •% 

X 

^ 

JO' 

•2/ 

' 

\ 

y2 

=  sin  30°. 

S 

s^ 

/ 

\ 

tan  60°  —  A/3 

X 

^ 

1 

- 

/ 

"t 

V 

V 

2" 

^ 

\ 

=  cot  oO  . 

S 

x^ 

1 

] 

tin  30° 

Functions  of  60° 
Equilateral  triangle. 

A/3                   Functions  of  30° 
=  cot  60°.       Equilateral  triangle. 

V3 


V3 


These  diagrams  should  be  memorized  as  half  of  a  unit 
square  for  45°,  and  half  of  an  equilateral  triangle  placed  ver- 
tically for  the  functions  of  60°  and  directly  related  angles, 
and  the  same  placed  horizontally  for  the  functions  of  30°  and 
related  angles  (-30°,  150°,  210°). 

sinO°  =  0.  sin  90°  =  1. 

cos  0°  =  1.  cos  90°  =  0. 

tanO°  =  0.  tan  90°  =  00. 


124 


The  meaning  of  the  expression  tan  90°  =  oo  (infinity)  is  that 
as  the  angle  0  approaches  nearer  to  90°  the  tangent  becomes 
larger  than  any  quantity  we  may  assign,  however  large; 
strictly  at  90°  the  tangent  function  has  no  meaning,  as  a  divi- 
sion by  zero  is  involved.  The  expression  tan  90°  =  x  is  not, 
then,  an  equality,  like  tan  60°  =  V3. 


Construction  of  the  regular  decagon 

OM  divides  OA  in  "  extreme  and  mean  "  ratio. 

Algebraical  method  by  solving,  3?  —  10  (10  —  x). 

The  method  of  constructing  a  decagon  combined  with  the 
solution  of  a  quadratic  equation  enables  us  to  find  the  sine  of 
18°.  The  radius  of  the  circle  is  divided  in  extreme  and  mean 
ratio  to  obtain  the  side  of  the  inscribed  decagon :  10(10— x)=  x*, 
in  a  circle  of  radius  10.  Whence, 

z2  +  10  x  -  100  =  0, 

x  =  -  5  ±  V125  =  -  5  ±  11.1803  =  -  16.180  or  +  6.1803, 
of  which  we  take  the  positive  value.     One  half  of  this  value 
is  the  value  of  y  in  the  triangle  of   reference  for  18°  when 
r  =  10.     Hence  the  sine  of  18°  is 
3.090 


10 


=  .3090. 


TRIGONOMETRIC  FUNCTIONS 


125 


9.  The  Greek  method,  using  chords.  —  By  the  methods  of  plane 
geometry,  using  chords  instead  of  half-chords,  the  sine  of  half  an  angle 
and  the  sine  of  the  sum  and  the  difference  of  the  two  given  angles  can  be 
computed.      One    theorem  involved,   in    addition    to  the  Pythagorean 
theorem,  is  not  given  in  many  geometries.     It  is  called  Ptolemy's  the- 
orem, as  it  is  fundamental   in  ,  the   method   of   computing   chords  de- 
veloped by  Ptolemy,  a  Greek  writer  of  the  second  century  A.D.,  whose 
textbook  on  astronomy,  the  Almagest,  continued  in  active  use  for  fifteen 
hundred  years.     The  theorem  is  that  in   an  inscribed  quadrilateral  the 
product  of  the  diagonals  is  equal  to  the  sum  of  the  products  of  the  oppo- 
site sides. 

From  the  chord  of  60°  one  can  compute  the  chord  of  30°  ;  thus  the 
sine  of  15°  is  obtained.  From  36°  and  30°  the  sine  of  3°  can  be  obtained 
by  using  hah'  the  chord  of  the  difference  of  two  given  arcs  ;  from  this  the 
sine  of  lj°,  f°,  |°,  ^6°,  3V,  ,y,  T|j°,  5|s°,  ...  can  be  computed.  The 
sine  of  1°  cannot  be  obtained  by  this  process,  nor  can  the  sine  of  £°  ; 
these  are  found  by  other  methods,  giving  approximations  as  accurate  as 
desired  for  any  practical  purposes. 

10.  Origin  of  the  tangent  and  cotangent  functions.  —  In  the 
study  of  astronomy  the  angle  of  inclination  to  the  horizon  of 


ftiiis 


H 


v/) 


Arabic  shadow  function 
The  shadow  varies  as  the  cotangent  of  the  angle  of  inclination  of  the  sun. 

the  sun  and  of  other  heavenly  bodies  is  important.     The  ratio 
of  the  length  of  the  shadow  to  the  length  of  the  vertical  object 


126 


UNIFIED  MATHEMATICS 


casting  the  shadow  gives  the  cotangent  of  the  angle  of  in- 
clination of  the  sun.  This  function  of  the  angle  appeared 
before  the  tangent  function  in  the  works  of  the  Arabic  as- 
tronomer, Al-Battani,  of  the  tenth  century  A.D.,  and  it  was 
called  the  shadow  and  later,  right  shadow  or  second  shadow. 
The  tangent  function,  being  the  ratio  of  the  length  of  the 
shadow  cast  on  a  vertical  wall  to  the  length  of  a  stick  placed 
horizontally  out  from  the  wall,  was  called  later  the  first 
shadow.  The  Arabs  took  the  length  of  the  stick  as  12. 


Variation  of  sine  and  cosine  as  9  varies. 

11.  Variation.  —  As  0  varies  the  trigonometric  functions 
also  vary ;  it  is  desirable  to  fix  in  mind  the  changes  of  the 
three  principal  functions,  viz.  sin  0,  cos  0,  and  tan  0,  as  0 
changes  by  rotation  of  the  moving  arm. 

Taking  r  =  10,  it  is  an  easy  matter  to  follow  on  the  graph 


the  changes  in  the  x  and  y  of  the  end  of  the  moving  ray.  As 
the  moving  ray  starts  from  OX,  an  angle  of  0°,  the  y  or  ordi- 
nate is  zero.  So  we  have  that  the  sine,  -,  begins  at  zero  for 

r 

0° ;  as  0  increases  the  y  increases,  reaching  a  maximum  of  10 
when  6  is  90°  and  the  maximum  value  then  of  sin  0  is  \$  or  1. 
As  0  increases  beyond  90°,  the  ordinate  begins  to  decrease, 
arriving  finally  at  0  when  the  moving  arm  is  on  OX'.  For 
angles  greater  than  180°  up  to  270°  the  ordinate  decreases, 
finally  reaching  a  minimum  or  lowest  value  of  —  10 ;  the  cor- 
responding minimum  of  sin  0  is  —  1 ;  from  270°  on  to  360°, 
completing  a  revolution,  the  sine  increases  from  —  1  up  to  0. 

For  angles  greater  than  360°,  or  for  negative  angles,  the 
moving  ray  would  move  through  no  new  positions ;  for  any 
such  angle  the  trigonometric  functions  are  equal  to  the  func- 
tions of  the  corresponding  positive  angle  having  the  same 
position. 

The  limits  +  1  and  —  1  of  the  sine  function  and  cosine 
function  are  evident,  of  course,  in  the  figure.  In  any  position 
of  the  moving  ray  x  and  y  are  the  sides  of  a  right  triangle  of 
which  r  is  the  hypotenuse,  except  that  on  the  axes  x  or  y 

equals  r ;  hence  the  quotients  -  and  -  are  either  numerically 

less  than  1  or  at  most  equal  to  1. 

Note  particularly  on  the  diagram  the  sines  of  30°,  45°,  and 

60°,  as  A,  approximately  111,  and  ^;  the  values,  .500, 0.707, 

and  0.866  may  well  be  memorized.  On  the  diagram  it  is  a 
simple  matter  to  read  the  sines  of  10°,  20°,  30°,  40°,  50°,  60°, 
70°,  80°,  and  90°  as  the  corresponding  ordinates  divided  by  10, 
correct  to  two  decimal  places.  The  cosines  of  these  angles  are 
read  as  the  corresponding  abscissas  divided  by  10. 

The  tangent  as  -  is  not  in  a  form  to  give  the  numerical  value 

without  computation ;  however,  by  drawing  the  tangent  line 
to  the  circle  at  A  and  producing  r  to  cut  the  tangent  line  at 


128 


UNIFIED  MATHEMATICS 


A  T     v  AT 

T,  you  have  =—  =  *- ;  whence  -  —  =  tan  a,  and  so  the  value  of 
OA     x  10 

the  tangent  of  the  angle  can  be  read  as  the  ordinate  at  A  di- 
vided by  10. 


The  tangent  read  as  a  length 
A  T  =  10  tan  «. 


tan  30°= -^  =  ^=.58; 
V3      10 

tan  45°  =  ^  =  !; 

-./Q       -17  q 

tan  60°  =  ^  =  ^  =  1.73. 

When  the  angle  increases 
beyond  90°  the  position  of  the 
terminal  arm  fixes  the  sign 
of  each  function ;  the  sine  is 
positive  when  the  arm  is  in 
the  upper  quadrants,  I  and 
II,  and  negative  in  the  lower, 
and  the  cosine  positive  to  the 

right,  I  and  IV,  and  negative  in  II  and  III.  The  tangent  is 
positive  in  I  and  III,  and  negative  in  II  and  IV ;  when  posi- 
tive the  corresponding  vertical  lengths  are  cut  off  above  A  on 
the  tangent  at  A,  and  when  negative,  in  II  and  IV,  the  corre- 
sponding vertical  lengths  are  cut  off  below  A  on  the  tangent. 

If  the  radius  is  taken  as  unity,  the  ordinate,  abscissa,  and 
tangent  length  represent  numerically  and  in  algebraic  sign  the 
sine,  cosine,  and  tangent  values  of  the  corresponding  angle. 
However  it  is  usually  more  convenient  to  take  a  radius  of  10, 
25,  50,  or  100  and  to  interpret  the  trigonometric  functions  as 
ratios,  as  indeed  they  are. 

12.  Related  angles.  —  From  our  definitions  it  is  evident  that 
sin  0  has  the  same  value  for  two  angles,  symmetrically  placed 
with  reference  to  the  y-axis,  6  and  180°  —  6 ;  cos  0  has  the 
same  value  for  two  angles  symmetrically  placed  with  respect 
to  the  x-axis,  6  and  —  0,  or  0  and  360°  —  0 ;  tan  0  has  the  same 


TRIGONOMETRIC  FUNCTIONS 


129 


value  for  two  angles  which  differ  by  180°,  6  and  180°  +  6.  All 
functions  are  the  same  for  angles  which  differ  by  360°,  or  by 
any  integral  (positive  or  negative)  multiple  of  360°,  for  the 
terminal  arms  of  such  angles  will  coincide  when  the  angles  are 
placed  in  position  to  determine  the  trigonometric  functions. 

The  trigonometric  functions  of  360°  -  6,  180°  -  6,  180°  +  0, 
90°  +  6,  90°  —  6,  and  —  0,  in  terms  of  the  functions  of  0  are  of 
particular  importance  in  later  work.  In  the  figure  the  vectors 


I 


a 


f 


v 


0,  -9,  180°  -0,  180°  +  0.     Related  angles 
Read  the  corresponding  functions  on  the  diagram. 

OPi,  OP2,  OP3,  and  OP4  are  the  terminal  arms  of  related 
angles  in  quadrants  I,.  II,  III,  and  IV.  The  vector  OPi  de- 
termines, we  may  say,  a  positive  acute  angle  XOPi,  and, 
further,  any  angles  which  differ  from  the  positive  acute  angle 
by  any  integral  multiple  of  360° ;  OP2  represents  the  terminal 
arm  of  180°-  0,  OP3  of  180°+  6,  and  OP4  of  360° -0,  or  of  -  9. 
If  0  is  the  angle  represented  in  quadrant  1, 180°  —  d  is  the 
angle  here  represented  in  II ;  and  conversely,  if  0  is  in  II, 
180°-  $  is  in  I ;  if  0  is  the  angle  in  III,  180°-  0  is  the  angle  in 
IV,  and  conversely.  Evidently  if  d  in  I  is  30°,  180°  -  6  is  150°, 
represented  in  II,  and  if  0  =  150°,  180°  -  6  =  30° ;  further,  if  Q  is 
-  330°  in  I,  differing  from  30°  by  -  360°,  180°  -  0  will  be 
180°  -(-  330°)  or  510°  which  is  in  II,  360°  +  150°,  differing 
from  180°-  30°  by  360°.  The  ordinates  in  I  and  II  are  equal 


130  UNIFIED   MATHEMATICS 

and  of  the  same  sign,  and  similarly  the  ordinates  in  III  and  IV 
are  algebraically  equal  ;  r  is  the  same  in  all.     Hence  for  all 

angles  0, 

sin  (180"  -  9  1  =  sin  9. 

The  abscissas  in  I  and  II  are  numerically  equal  but  opposite 
in  sign,  similarly  in  III  and  IV  :  the  vectors  are  the  same  and 
positive. 

Hence  cos  (180°  -  6)  =  -  cos  6. 


By  definition,  tan  (180°  -  6)  =  ,  for  all  angles. 

cos  (18v  —  6) 

Bv  substitution, 


this  formula  also  holds  for  all  angles  0,  since  every  formula  in- 
volved has  been  shown  to  hold  for  all  angles  6. 

If  0  is  an  angle  in  I,  180°  4-  6  is  in  III  ;  the  corresponding 
positions  are  represented  for  any  such  angles  by  our  figure. 
If  0  is  represented  by  the  vector  in  II,  180°  +  6  is  represented 
by  the  vector  in  IV.  The  ordinate  in  III  equals  numerically 
the  ordinate  in  I,  but  is  opposite  in  sign  ;  similarly  the  ab- 
scissas of  I  and  III  ;  similarly  the  ordinates  and  abscissas, 
respectively,  of  II  and  IV  are  equal  in  value  and  opposite  in 
sign. 

Hence  sin  <180"  +  9)=  —  sin  6. 

and  cos  <  180°  +  0)=  —  cos  6. 

These  equalities  hold  for  all  angles  0. 


By  definition,  tan  (180°+g)  =  o=  =  tan  0, 

cos(180°+0)     —  costf 

which  holds  for  all  angles  B. 
In  precisely  the  same  way 

sin  (—  9)  =  —  sin  9. 
cos  (-  8)=  cos  9, 
and  tan  (—  6)  =  —  tan  9,  for  all  values  of  0. 


TRIGONOMETRIC  FUNCTIONS 


131 


e  ,  90°  +  6.     Related  angles  which  differ  by  90° 
Our  second  figure  can  be  used  to  show  that 

sin  (90°  +  0)=  cos  0, 
cos  (90°  +  0)=-  sin0, 

for  all  values  of  6.  If  6  is  the  angle  represented  in  I,  90°  +  6 
is  the  angle  here  represented  in  II  ;  if  0  is  in  II,  90°  4-  0  is  rep- 
resented in  III  ;  if  d  is  in  any  quadrant,  90°+  6  is»m  the  quadrant 
following  in  the  counter-clockwise  sense.  Now  the  ordinate  in 
any  quadrant  here  equals  numerically  and  algebraically  the 
preceding  abscissa  ;  thus  y2  =  x1;  y3  =  x.2;  y^  =  x3  ;  y^  =  x±. 

For  any  angle  6,  sin  (90°  +  6)  =  cos  8. 
Similarly  cos  (90°  +  6)  =  —  sin  0, 

tan  (90°  +  ° 


cos  (90°  +  0)      -  sin  6 
=  —  cot  6. 


132  UNIFIED  MATHEMATICS 

The  following  relations  have  now  been  established  for  all 
angles  6: 

sin  (180°  -  6)  =  sin  9,  sin  (180°  +  9)  =  -  sin  6, 

cos  (180°  -  6)  =  -  cos  6,  cos  (180°  +  6)  =  -  cos  6, 

tan  (180°  -  6)  =  -  tan  9,  tan  (180°  +  6)  =  tan  9, 

sin  (—9)=—  sin  9, 
cos  (—9)=  cos  9, 

tan  (—9)  =  tan  9. 
The  formulas, 

sin  (90°-  9)=  cos  9, 

cos  (90°  -  9)  =  sin  9, 

and  tan  (90°  -  9)  =  cot  9, 

have  been  established  for  acute  angles.  However,  the  preced- 
ing formulas  for  90°  +  0  which  we  have  established  for  all 
values  of  0,  positive  and  negative,  can  be  used  to  prove  that 
these  formulas  for  90°  —  B  hold  for  all  values  of  6. 

Thus  sin  (90°  —  6)  can  be  considered  as  sin  (90°  +  (  —  0)), 
and  as  the  formulas  for  90°  +  0  hold  for  all  values  of  6,  we 
have: 

sin  (90°  +  (—  0))=  cos  (-  0)=  cos  6, 

and  cos  (90°-f  (-  0))  =  —  sin  (_#)  =  _ (_  sin0)=  +  sin0. 

Hence  sin  (90°  -  6}  =  cos  0, 

and  cos  (90°  —  0)  =  sin  0,  for  all  values  of  0. 

T?     . ,  /nno      /IN      sin  (90°  —  0}      cos  6 

Further  tan  (90  —  6)  =  -  -1  =  —  —  =  cot  8, 

cos  (90°-  0)      sin  0 

again  for  all  values  of  0. 

The  student  will  do  well  to  remember  the  diagrams  and  to 
connect  the  formulas  with  these.  It  is  necessary  to  recollect 
only  the  representation  for  an  acute  angle  0 ;  it  is  more  desir- 
able to  connect  the  formulas  with  the  diagrams  than  merely  to 
memorize  the  formulas. 


TRIGONOMETRIC  FUNCTIONS  133 

PROBLEMS 

1.  Find  the  sine,  cosine,  and  tangent  of  210°.     The  triangle 
is  the  same  as  that  used  for  the  functions  of  30°,  but  it  is 
placed  in  III. 

2.  sin  150°  =         ;  cos    150°  =         ;  tan    150°  = 

3.  sin  315°  =         ;  cos  -  45°  =         ;  tan  -  45°  = 

4.  sin  225°  =         ;  sin    495°  =         ;  tan    750°  = 

5.  Express  the  following  in  terms  of  functions  of  positive 
angles  less  than  45° : 

a.    sin  170°=  b.   sin     130°  = 

c.   cos  170°  =  d.   cos    130°  = 

e.   sin  220°  =  /.    tan  -  40°  = 

6.  Express  in  terms  of  functions  of  x : 
.    a.    sin  (x- 90°)  = 

HINT.  —  Use  first  sin  (  —  0)  =  —  sin  6. 

b.  sin  (270°  +  a)  = 

HINT.  —  Express  first  as 

(180°  +  6) ;    i.e.  sin  (180°  +  90°  +  z)  =  -  sin  (90°  +  z)  =  .... 

c.  cos  (x-  270°)=  d.   tan (360°-  a;)  = 

7.  Draw  one  quadrant  of  a  circle  of  radius  10  half-inches ; 
construct  the  angles  of  30°,  45°,  and  60°  and  read  their  values. 
Bisect  the  angle  of  30°  and  so  obtain  the  values  of  the  func- 
tions of  15°.     Make  a  table  of  values  of  sines,  cosines,  and  tan- 
gents, advancing  by  15°.      Note  that  the  chord   of  30°  may 
readily  be  computed ;  one  half  of  this  chord  divided  by  the 
radius  gives  the  sine  of  15°.     Find  the  sine  of  7^°  similarly. 
From  the  table   of  sines   of  acute   angles  from  0  to  90°  by 
15°  intervals,  give  the  sines  of  the  related  obtuse  angles  up 
to  180°. 

8.  Given  tan  0  =  3,  find  sec  0  and  cos  0 ;  what  is  the  signifi- 
cance of  the  double  sign  in  the  answer  ? 


134 


UNIFIED   MATHEMATICS 


45° 


9.   Express  in  terms  of  functions  of  positive  angles  less  than 


a.  sin  100°. 

b.  cos  100°. 

c.  tan  100°, 

d.  sin  200°. 


e.  tan  200°. 

/.  sin  300°. 

g.  cos  (—60°). 

h.  cos  (-160°). 


t.  tan  (-420°). 

j.  sin  750°  50'. 

k.  cos  1030°  40'. 

I  tan  218°  10'. 


13.    Angles  constructed   from   given   functions.  —  Given 


sin  0  =  4, 


construct  both  values  of  0  (in  I  and  II).     The  problem  is 
in  geometrical  language  to  construct  a  right  triangle  with  the 

hypotenuse  and  one  side  given,  since  sin0  =  ^. 

r 

Take  r  as  7  and  y  as  4  ;  since  sin  0  =  - ,  7  is  to  be  one  side  of 

r 
our  angle  ;  with  7  as  a  radius  and  0  as  center  describe  a  semi- 


Graphical  solution  of  sin  6  =  - 
7 

circle  above  the  cc-axis  ;  y  =  4  is  a  line  parallel  to  the  axis  of  x, 
cutting  the  circle,  x-  +  y2  =  49,  in  two  points.  Find  the  inter- 
sections ax  and  a2 ;  geometrically  the  angle  is  found.  Using 
the  Pythagorean  theorem  a2  -f  42  =  72,  and  xz  =  33,  x  =  ±  5.75. 
The  positive  value  of  x  is  to  the  right  and  the  negative  to  the 


TRIGONOMETRIC   FUNCTIONS 


135 


left;  to  the  first  corresponds   the  acute  angle  0,  and  to  the 
other  0  with  the  terminal  arm  in  II. 

In  I,  sin  6  =  ^  =  .57.  In  II,  sin  0  =  |. 

5.75 


cos  e  =   ™  =  .82. 

7 

tan  B  —  -'—=  .70. 
5.75 


cos    =  -—  =  - 


tan  6  =  — 


7 
4 


5.75 
=  -  .70. 

This  problem  should  also  be  solved  using  the  formula 

sin2  6  +  cos2  0  =  1. 

Given  tan  0  =  1.4,  find  the  other 
functions  of  0  and  discuss  the  two 
solutions. 

cos  6     x 
0  can  be  in  I  or  III. 

Take  y  as  1.4,  a;  as  1  (or  y  as 
14,  x  =  10,  or  other  values  as  con- 
venient) ;  evidently  x  as  —  1  and 
y  as  -  1.4  gives  0  in  III.  Graphical  solution  of  tan  t  =  1.4 

Since  x2  +  y*  =  r2,  ?-2  =  1  -f  1.96  =  2.96. 

r  =  1.72+ 

In  I,  sin0  =  ^   -   or   -M_  =  M.  V2T96  =  .81  (or  .814  to 
l.<2         V2.96     2.96 

three  places). 

cos0  =  -      =.581+. 
1.72 

In  III,  sin  0  =  -  .814, 
cos  0  =  —  .581. 

This  problem  should  be  solved  also  by  using  the  formula 
1  -f  tan2  6  =  sec2  0. 


136  UNIFIED  MATHEMATICS 

EXERCISES 

1.  Use  10  of  the  larger  units  on  a  sheet  of  cross-section 
paper  and  find  by  construction  the  sines  of  the  angles  0°,  10°, 
20°,  30°,  .»  180°.      Compare  with  the  table.      Note  that  the 
values  for  10°  and  170°,  20°  and  160°,  30°  and  150°  •••  corre- 
spond.     Use   the   formula  cos  0  =  sin  (90°  —  0)   to   find   the 
cosines  of  0°,  10°,  20°,  —  90°.    Note  that  in  the  second  quadrant 
the  cosines  become  negative. 

2.  Construct  an  angle  of  60°,  using  10  as  side  of  the  equi- 
lateral triangle  used.     Find  cos  60°,  sin  60°,  tan  60°  to  2  places. 

3.  Find  the  sine  of  150°,  210°,  330°. 

Use  half  the  equilateral  triangle,  placed  horizontally  with  vertex  of  30° 
angle  placed  at  the  origin. 

4.  Find  the  sine  and  cosine  of  120°,  135°,  225°,  -  30°. 

5.  Find  the  tangent  of  120°,  135°,  225°,  -  30°,  from  the 
data  of  the  preceding  problem ;  find  tan  120°,  tan  135°,  tan  225°, 
tan  (—  30°)  from  the  geometrical  figure. 

6.  By  construction  of  a  square,  side  10  units,  find  approxi- 
mate values  of  the  functions  of  45°.     Find  the  values  using  the 
Pythagorean  theorem. 

7.  Construct    an  angle  of    30°,   and   find    values    of    the 
functions. 

8.  Construct  angles  of  15°  and  7|-0,  and  find  the  values  of 
the  functions. 

9.  Given  sin  0  =  -fa,  find  cos  0  and   tan  0 ;    indicate  both 
solutions. 

10.  Given  cos  6  =  .432,  compute  sin  6  and  tan  0  to  3  places, 
6  in  I. 

11.  Given  tan  0  =  4.32,  compute  sin  0  and  cos  0  for  0  in  III. 

12.  Given  tan  0  =  \ ,  construct  0  geometrically. 

13.  Construct  9  geometrically,  given  sin  0  =  T^-. 

14.  Given  sin  0  =  —  1AS-,  find  cos  6,  0  in  IV. 

15.  Given  sin  0  —  .43,  find  cos  0,  0  in  III. 


TRIGONOMETRIC  FUNCTIONS  137 

16.  Given  tan  9  =  —  .43,  construct  0  in  II,  and  find  values 
of  sin  0  and  cos  0  from  the  figure. 

14.  The  inverse  functions.  —  If  we  are  given  the  sine  s  of  an 
angle  and  desire  to  speak  of  the  angle  we  can  say  "the  angle 
whose  sine  is  s"  and  we  can  abbreviate  this  expression  in 
writing  to  arc  sin  s  or  to  sin"1  s.  Similarly  the  angle  whose 
cosine  is  ra  is  written  arc  cos  m,  or  cos"1  m.  Note  that  in  sin"1  s, 
cos"1  m,  and  tan"1  k,  the  —  1  is  not  at  all  a  negative  expo- 
nent ;  these  expressions  for  angles  are  read  anti-sine  s,  anti- 
cosine  m,  and  anti-tangent  k,  or  sometimes,  inverse  sine  s,  etc., 
respectively. 

In  what  follows  we  shall  use  mainly  the  symbols  arc  sin, 
arc  cos,  arc  tan,  arc  esc,  arc  sec,  and  arc  cot,  although  the 
other  symbols  are  also  in  common  use.  Whether  the  "  arc  "  or 
"  —  1  "  symbols  are  used  the  student  is  strongly  advised  to 
read  "  arc  tan  t "  or  "  tan"1 1 "  always  as  "  the  angle  whose  tan- 
gent is  t"  and  similarly  expressions  like  arc  cos  cc,  arc  sin  1,  and 
sin"1  k. 

A  given  angle  has  only  one  sine,  but  a  given  number  is  the 
sine  of  many  different  angles.  A  similar  remark  applies  to 
the  other  five  functions.  To  illustrate :  sin  30°  is  0.5  and  no 
other  value.  But  arc  sin  .05,  the  angle  whose  sine  is  0.5,  may 
be  30°  or  -  330°  or  390°  or  750°  or  150°  or  510°  or  870°  or  any 
angle  differing  from  30°  or  150°  by  an  integral  multiple  of  360°. 
The  sine  of  any  one  of  these  various  angles  is  0.5 ; 

sin  (k  360°  +  30°)  =  .5  and  sin  (k  360°  -f  150°)  =  .5, 
where  k  is  any  integer. 

PROBLEMS 

1.  Given  arc  cos£  =6,  construct  0  both  in  the  first  and  in 
the  fourth  quadrant.     Note  that  the  problem  is  precisely  the 
same -as  though  the  requirement  were  to  construct  0  when 
given  that  cos  0  =  -J  ; .  or  to  construct  arc  cos  ^. 

2.  Between  what  values  must  k  lie  to  have  any  solution  for 
0  =  arc  cos  k  ? 


138 


UNIFIED  MATHEMATICS 


3.  Given  that  the  angle,  arc  sin  ^,  is  obtuse,  construct  the 
angle. 

4.  Construct  the  following  angles  of  the  first  quadrant :  arc 
sin^,  arc  tan  (+  2),  arc  cos  y^,  arc  sin  .43.     Give  the  approxi- 
mate value  of  the  other  two  principal  functions  in  each  case. 

5.  Give  five  solutions  of  each  of  the  following : 

arc  cos  ^  =      ;  arc  tan  1  =      ;     arc  sin  0  =      ;     arc  cos  1  = 

6.  If  arc  sin  .438  =  26°,  what  is  arc  cos  .438  ? 

7.  What  is  the  value  of  arc  sin  —  .438?    Give  four  answers. 
Give  the  general  formula  representing  angles  9  which  satisfy 
B  =  arc  sin  (-  .438).     What  is  arc  cos  -  .438  ? 

8.  On  the  following  diagram,  regarding  the  circle  as  having 
a   radius   of   100,  read   the  numerical    value  to   two  decimal 


places,  of  the  sine,  cosine,  and  tangent  of  each  angle  repre- 
sented.    Each  minor  division  represents  4  units. 


CHAPTER   VIII 

TABLES   AND   APPLICATIONS 

1.  Tables.  —  The  tables  of  the  trigonometric  functions  are 
computed  by  processes  dependent  upon  formulas  derived  in 
the  higher  mathematics.  We  have  shown  the  graphical 
method  of  finding  sine,  cosine,  and  tangent,  which  serves  also 
to  bring  out  the  fact  that  the  sines  of  angles  from  0  to  45°  are 
at  the  same  time  cosines  of  the  complementary  angles ;  simi- 
larly since  tan  (90°  —  x)  =  cot  x,  it  follows  that  the  tangents  of 
angles  from  0  to  45°  are  cotangents  of  the  complementary 
angles,  from  90°  down  to  45°.  Since  tables  are  given  of  both 
sine  and  cosine  it  is  necessary  to  give  values  of  both  functions 
only  up  to  45°,  and  similarly  with  tangent  and  cotangent. 
Thus  sin  26°  10'  is  found  in  the  table,  of  sines  which  reads 
down  with  26°  at  the  left,  and  below  10'  as  given  at  the  top ; 
if  the  cos  63°  50'  is  sought  we  look  for  63°  at  the  right  of  the 
table  of  sines  with  the  minutes  to  be  read  below ;  and  we  find 
that  the  cosine  table  is  the  same  as  the  table  of  sines,  but 
reading  up ;  this  brings  us  to  precisely  the  same  place  in 
the  tables  as  sin  26°  10',  the  complementary  angle ;  similarly 
sin  63°  50'  is  sought  in  the  row  marked  63  at  the  left  of  the 
table  and  leads  to  the  value  which  read  as  a  cosine  represents 
cos  26°  10'. 

For  angles  greater  than  90°  the  formulas  which  we  have 
given  for  related  angles  are  applied.  Probably  the  simplest 
formulas  to  apply  to  obtain  the  functions  of  obtuse  angles  are 
the  formulas, 

sin  (90°  +  x)  =  cos  x, 

and  cos  (90°  +  a;)  =  —  sin  x. 

139 

\ 


140  UNIFIED  MATHEMATICS 

Thus  sin  128°  35'  =  cos  38°  35' ; 

cos  128°  35'  =  -  sin  38°  35'. 

It  is  well  to  note  that  subtracting  90°  from  angles  greater 
than  100°  and  less  than  200°  simply  increases  the  tens'  digit 
of  the  angular  measure  by  one,  dropping  the  hundreds'  digit. 

The  formulas  for  180°  +  a,  and  for  360°  —  «  or  —  «,  are 
used  for  angles  in  III  or  IV. 

Since  computation  is  largely  effected  by  means  of  logarithms, 
it  becomes  desirable  to  have  separate  tables  of  the  logarithms 
of  the  trigonometric  functions.  The  sines  and  cosines  of  all 
angles  are  numerically  less  than  1  and  so  are  tangents  of 
angles  less  than  4t>° ;  hence  the  logarithms  of  these  numbers 
will  have  negative  characteristics.  In  the  logarithms  of  the 
trigonometric  functions,  —  10  is  to  be  annexed  to  the  logarithm 
as  given  in  the  table  for  sines,  cosines,  and  tangents  up  to  45°. 
Thus  log  sin  30°  is  9.6990  -  10  ;  log  sin  56°  10'  =  9.9194  -  10 ; 
log  tan  34°  20'  =  9.8317  -  10 ;  but  log  tan  56°  10'  =  .1737. 

2.  Interpolation.  —  The  insertion,  by  interpolation,  of  the 
natural  and  logarithmic  functions  of  angles  lying  between 
those  expressly  given  in  the  tables  follows  precisely  the  same 
lines  as  in  the  corresponding  problem  in  the  logarithms  of 
numbers.  Our  tables  give  these  functions  for  angles  increas- 
ing by  multiples  of  10  minutes  ;  interpolation  enables  us  to 
compute  the  functions  of  angles  to  minutes ;  in  using  tables 
giving  the  functions  to  minutes  interpolation  enables  us  to 
compute  to  tenths  of  a  minute.  Note  that  the  assumption  is 
always  that  if  angles  are  read  to  minutes  you  compute  only  to 
minutes ;  the  tables  used  should  correspond  to  the  precision 
of  measurement  of  the  given  data.  Four-place  tables  are, 
in  general,  sufficiently  accurate  for  measurements  which  are 
made  to  four  places  in  numbers,  and  to  minutes  in  angular 
measurement. 

Illustrative  problems.  1.  Find  by  interpolation  (a)  sin  36°  15', 
(6)  log  sin  36°  16',  (c)  log  cos  36°  18',  and  (d)  log  tan  36°  14'. 


TABLES  AND  APPLICATIONS  141 

TABULAR  VALUES — Compare  with  your  tables 


angle 

sin 

log  sin 

COS 

log  cos 

log  tan 

36°  10' 

.5901 

97710 

.8073 

9.9070 

9.8639 

53°  50' 

36°  20' 

.5925 

9.7727 

.8056 

9.9061 

9.8666 

53°  40' 

cos 

log  cos 

sin 

log  sin 

log  cot 

angle 

The  values  to  four  decimal  places  of  the  functions  of  angles 
between  36°  10'  and  36°  20'  evidently  lie  between  the  values 
which  are  here  given.  Thus  sin  36°  10'  is  .5901  and  sin  36°  20' 
is  .5925,  an  increase  of  24  units  of  the  fourth  place ;  this  24 
is  called  the  tabular  difference.  To  the  ten  equal  steps  of  in- 
crease from  36°  10'  to  36°  20',  by  minutes,  correspond  ten  in- 
creases approximately  equal  to  each  other,  in  the  sines  of  these 
angles,  making  a  total  increase  in  ten  steps  of  24  units  of  the 
fourth  place.  The  tenths  of  24  are  respectively, 

.1         .2         .3         .4  .5  .6  .7  .8         .9 

2.4       4.8       7.2       9.6  12       14.4       16.8        19.2      21.6 

In  adding,  as  our  logarithms  are  given  only  to  four  places, 
we  add  rejecting  tenths,  and  retaining  in  the  last  place  the 
nearest  unit.  Thus  for  tenths  of  24  we  use  always  2,  5,  7,  10, 
12,  14,  17,  19,  and  22.  The  interpolation  does  not  always  give 
the  correct  result  to  four  places,  although  in  the  values  of  the 
sine  the  error  is  always  less  than  1  unit  of  the  fourth  place. 
In  the  above  values  of  sin  36°  11'  to  36°  19'  as  given  by  the 
addition  2,  5,  7,  10,  12,  14,  17,  19,  and  22  units  of  the  fourth 
place  to  .5701  the  first  value  .5703  should  be,  to  4  places, 
.5704 ;  the  error  is  less  here  than  -^  of  1  %  of  the  value  taken. 

1.   a.  sin  36°  15'  =  .5701  +  T%  of  .0024  =  .5713. 

Method :  Tabular  difference  is  24 ;  to  .5701  add  .5  of  24 
units  of  the  fourth  place. 

6.   log  sin  36°  16'  =  9.7710  - 10  +  T%(.0017)=  9.7720— 10. 
Tabular  difference  is  17 ;  10.2  is  replaced  by  10,  and  this  is 
added  in  the  third  and  fourth  decimal  places  to  9.7710. 


142  UNIFIED  MATHEMATICS 

0.  log  cos  36°  18'  =  9.9070  -  10  -  ^(.0009)  =  9.9063  -  10. 
Tabular  difference  is  9 ;  cosine  and  log  cosine  are  decreasing 

functions ;  7  units  of  the  fourth  place  must  be  subtracted. 

d.   log  tan  36°  14'  =  9.8639  - 10  +  T4ff(.0027)  =  9.8650  -  10. 
Tabular  difference  is  27  ;  10.8  is  replaced  by  11. 

2.  Find  to  minutes,  by  interpolating,  the  angle  when  given 
(a)  sin  a  =  .5919,  (6)  log  sin  a  =  9.7717,  and  (c)  log  cot «  = 

9.8650 ;  find  a  in  each  case. 

3.  (a)  sin  a  =  .5919 ;  tabular  difference  is  24 ;  given  differ- 
ence .5901  to  .5919  is  18  units  of  the  fourth  place.    Among  the 
tenths  of  24  find  the  nearest  to  18 ;  16.8  and  19.2,  respectively 
.7  and  .8  of  24,  are  equally  near  and    the   even  number  of 
tenths  is  commonly  taken,  in  such  cases,  by  computers. 

sin  a  =  .5919  ;  a  =  36°  18'. 

(6)  log  sin  a  =  9.7717  ;  a  =  36°  10'  +  T?r  of  10'  (to  minutes). 

a  =  36°  14'. 

Tabular  difference  is  17  ;  7  is  nearest  to  .4  of  17. 
(c)  log  cot  a  =  9.8650  ;  a  =  53°  40'  +  |f  of  10'. 
Tabular  difference  is  27,  a  decrease;  given  decrease  is  16; 
among  the  tenths  of  27  the  nearest  to  16  is  6 ;  hence  a  =  53°  46'. 
Had  log  cot «  been  given  as  9.8651  -  10  or  9.8649  — 10,  the 
angle  a  would  again  be  given  as  53°  46'. 

PROBLEMS 

1.  Find  the  20  natural  trigonometric   functions  following, 
without  interpolation ;  time  yourself ;  limit  6  minutes. 

a.  sin  36°  10'.  g.  tan  70°  30'. 

b.  tan  63°  20'.  h.  sin  28°  50'. 

c.  cos  34°  10'.  t.  tan  16°  20'. 

d.  cot  80°  00'.  j.  cos  8°  40'. 

e.  sin  59°  30'.  k.  sin  157°  10'. 
/.  cos  48°  50'.  I  cos  214°  10'. 


TABLES  AND  APPLICATIONS 


143 


m.  cot  141°  00'. 

n.  tan  329°  30'. 

o.  cos  136°  50'. 

p.  cos  -28°  10'. 


q.  tan-64°20'. 

r.  sin  384°  00'. 

s.  cot  756°  00'. 

t.  sin  242°  40'. 


2.   Find  the  logarithms  of  the  above  20  trigonometric  func- 
tions, timing  yourself.     Limit  7  minutes. 


3.   Find   the   following  20 
yourself.     Limit  12  minutes. 


logarithms,  interpolating ;  time 


a. 

log 

sin  36° 

14'. 

/ 

log 

cos  48°  57'. 

Jc. 

log 

sin  152° 

15'. 

b. 

log 

tan  63° 

29'. 

9- 

log 

tan  70°  33'. 

I 

log 

cos  214° 

26'. 

c. 

log 

cos  34° 

14'. 

h. 

log 

sin  28°  51'. 

m. 

log 

cot  141° 

05'. 

d. 

log 

cot  80°  06'. 

i. 

log 

tan  16°  22'. 

n. 

log  tan  329° 

33'. 

e. 

log 

sin  59° 

32'. 

j- 

log 

cos  8°  48'. 

o. 

log 

cos  136° 

57'. 

P- 

log 

cos  -  28°  11'. 

Q- 

log 

tan  -  64°  26' 

r. 

log 

sin  384°  03'. 

s. 

log 

cot  756°  08'. 

t. 

log 

sin  242°  44'. 

4.   Find  the  angles  less  than  90°  corresponding  to  the  follow- 


ing 20  logarithms ;  no  interpolation ; 

a.  log  sin  a  =  9.6878  -  10  Jc. 

b.  log  cos  a  =  9.9954  -  10  I 

c.  log  tan  a  =  9.4858  -  10  m. 

d.  log  cot  a  =    .5102  n. 

e.  log  cos  a  =  9.8241  -  10  o. 

f.  log  tan  a  =  9.7873  -  10  p. 

g.  log  sin  a  =  9.3179  -  10  q. 
h.  log  tan  a  =    .2155  r. 
i.  log  cos  a  =  8.9816  -  10  s. 
>  log  cot  a  =  9.9341  -  10  t. 


time  6  minutes. 

log  sin  a  =  9.9499  -  10 

log  cos  a  =  9.8081  - 10 

log  cot  a  =    .8904 

log  tan  a  =  8.9420  -  10 

log  cos  a  =  9.9640  -  10 

log  cos  a  =  9.9757  -  10 

log  tan  a  =    .5720 

log  sin  a  =  8.9403  -  10 

log  cot  a  =    .0152 

log  sin  a  =  9.9977  -  10 


5.  Give  in  each  case  another  angle  which  would  satisfy  the 
above  relationship,  in  problem  4  ;  e.  q.  if  log  sin  a  =  9.6990—10, 
a  =  30°  or  150°. 


144 


UNIFIED  MATHEMATICS 


6.    Find  the  following  20  angles 


interpolate ; 
a.  log  sin  a 
6.  log  cos  a 

c.  log  tan  « 

d.  log  cot  a 

e.  log  cos  « 
/.  log  tan  a 
g.  log  sin  a 
h.  log  tan  a 

i.  log  cos  a 
.;.  log  cot  a 
A1,  log  sin  a 
/.  log  cos  a 
m.  log  cot  a 
w.  log  tan  a 
o.  log  cos  a 
/>.  log  cos  a 
g.  log  tan  a 
r.  log  sin  « 
s.  log  cot  a 
<.  log  sin  a 


time  yourself. 
=  9.6881  -  10 
=  9.9956  -  10 
=  9.4861  -  10 
=    .5104 
=  9.8228  -  10 
=  9.7879  -  10 
=  9.3200  -  10 
=    .2144 
=  8.9912  -  10 
=  9.9358  -  10 
=  9.9502  -  10 
=  9.8092  -  10 
=    .8955 
=  8.9492  -  10 
=  9.9645  -  10 
=  9.9753  -  10 
=    .5699 
=  8.9404  -  10 
=    .0137 
=  9.9978  - 10 


Angle  10°  in  a  circle  of  radius  5  inthes 

PM  =  .868  in. ;  arc  PA  =  .873  in.; 

AT-  .882  in. 


3.  Angles  near  0°  and  90°. 

—  For  angles  near  zero, 
from  0°  to  2°,  the  cosines 
vary  only  from  1.0000  to 
.9994;  the  cosine  function 
to  4  places  cannot  then  be 
used  for  determination  of 
the  angle  to  minutes.  Simi- 
larly, of  course,  the  sines 
of  angles  from  88°  to  90° 
vary  between  the  same 
limits.  For  ordinary  pur- 
poses it  will  suffice  to 
avoid  the  use  of  the  cosine 


TABLES  AND  APPLICATIONS 


145 


in  the  interval  from  0°  to  2°  or  3°  or  4° ;  the  method  of  avoid- 
ance is  explained  below. 

In  computing  graphically  the  values  of  sin  6  and  tan  0  even 
with  a  radius  of  10  cm.,  or  of  5  inches,  the  difference  between 
tan  6  and  sin  0  becomes  too  small  to  read  accurately  when  6 

is  less  than  —  (i.e.  7|°;  .131').     For  10°  which  is  .1745  radian, 

sin  .1745r  is  .1736  and  tan  0  is  .1763 ;  for  5°  or  .0873r, 
sin  #=.0872  and  tan0  is  .0875r;  for  1°  or  .Ol745r,  sin  0=. 01 745 
and  tan  0  is  .01746  or  5  places  are  necessary  to  exhibit  any 
difference  between  0,  sin  6,  and  tan  0. 


sin  0  <  0  <  tan  8 

for  small  acute  angles  0,  0  is  measured  in  radians 

Evidently,  triangular  area  GAP  <  sector  GAP  <  area  OAT, 
but  the  area  of  the  triangle 

OAP=±GAx  MP 

=  \  r  x  r  sin  Q 
=  £  r2  sin  B. 

The  area  of  the  sector  GAP  =  ^  r20,   since  0  is  measured  in 
radians,  and  the  area  GAT=^  r2  tan  0. 
Whence,  by  substituting, 

i r2 sin  6  <  l^B  <  ^f*  tan 0. 

e        i 


1< 


sin  0       cos  6 
8 


Whence,  as  6  diminishes,  -T— ,  lying  between  1  and  a  num- 
ber approaching  .1,  can  be  made  as  near  to  1  as  we  please.  By 
methods  of  plane  geometry,  using  30°,  15°,  7£°,  3|°,  together 


146 


UNIFIED  MATHEMATICS 


with  72°,  60°,  12°,  6°,  and  3°  it  can  be  established  that  cos  f ° 
differs  from  1  by  less  than  ^  of  1  %  ;  cos  £  °  =  .99991 ;  for 
any  angle  0,  less  than  f  °,  0  will  exceed  sin  0  by  less  than  T^  of 
1  °f0  and  tan  0  will  exceed  0  by  less  than  y^  of  1  %.  Similarly 
the  discrepancy  between  sin  6  and  tan  6  for  0,  any  angle 
less  than  3|°,  is  less  than  ^  of  1  %,  and 
for  any  angle  up  to  8°  the  difference  is 
less  than  1  %  of  either  value. 

On  the  earth's  surface  ordinary  dis- 
tances are  regarded  as  straight  lines. 
However  for  many  purposes  the  deviation 
from  a  straight  line  is  of  importance : 
thus  particularly  with  projectiles  of  long 
range,  the  deviation  is  of  vital  importance. 
In  the  figure  given  if  PA  represents  an 
arc  on  the  earth's  surface,  PT  may  be 
regarded  as  the  altitude  of  a  balloon, 
aeroplane,  or  top  of  a  mountain,  and  TA 
gives  the  distance  of  the  horizon.  /.  TO  A 
is  equal  to  the  dip  of  the  horizon.  AM 
is  the  drop  in  the  distance  twice  PA, 
i.e.  from  T  an  observer  would  note,  on 
the  ocean,  the  complete  disappearance  of  a 
ship  of  height  AM  when  the  ship  is  at  Q. 
By  algebraic  process,  A  T  =  V2  rh  +  hz ; 
when  h  is  measured  in  feet,  r  in  miles, 
and  TA  in  miles,  this  gives  for  values  of 

Jo    7  "1 

h  less  than  15  miles,  AT  =  A/ — -,  correct  to  -  of  1  %.     Check 
using  3960  miles  as  r. 


Arc  PAQ   on   the 
earth's  surface 

TA,  horizon  dis- 
tance. 

PT,  height  of  ob- 
server. 


PROBLEMS 


1.  Given  that  an  observer  is  at  a  height  of  1000  feet,  com- 
pute the  distance  to  the  horizon,  r  =  3960  miles.  What  is 
the  dip  of  the  horizon  ?  Note  that  the  tangent  of  the  dip- 
angle  is  the  horizon  distance  divided  by  the  radius. 


TABLES  AND  APPLICATIONS 


147 


2.  Find  the  angle  subtended  at  the  center  of  the  earth  by 
an  arc  of  length  1  mile,  10  miles,  20  miles. 

3.  What  is  1°  of  latitude  in  miles  ? 

4.  Degrees  of  longitude  vary  in  length  from  degrees  on  a 
great  circle  of   the  earth  at  the   equator  to  0  at  the  poles. 
Find  the  radius  of  the  small  circles  on  which  degrees  of  longi- 
tude are  measured,  for  40°  north  latitude.     Where  else  on  the 


^- 

L--  ~, 

X 

^ 

^ 

X 

/ 

/ 

•^. 

s 

/' 

X 

--" 

>cc 

.s4 

) 

^^ 

\A 

/ 

^~-, 

^_ 

^~ 

0-- 

s 

; 

.. 

/ 

\ 

/ 

1 

I 

n 

>   J 

/ 

\ 

t 

-y 

s 

\ 

/ 

s 

\ 

> 

L 

0' 

\ 

g 

/ 

i 

1 

Circle  of  40°  N.  latitude 

earth's  surface  would  degrees  of  longitude  be  the  same  ?  From 
35°  to  45°  N.  latitude  discuss  the  percentage  variation  in 
degrees  of  longitude,  as  compared  with  degrees  of  longitude 
at  40°  N.  latitude. 

5.  How  far  below  the  arc  of  1  mile  on  the  earth  does  the 
corresponding  chord  fall  at  the  lowest  point  ?     Find  the  same 
distance  in  inches  for  arcs  of  2  miles,  8  miles,  10  miles,  16 
miles,  20  miles. 

6.  What  part  of  the  height  of  a  mountain,  measured  on 
the  altitude,  is  not  visible  from  a  point  20  miles  distant  ? 

7.  From  what  distance  can  the  top  of  a  mountain  10,000 
feet  high  be  seen  ? 

8.  What    distance    from   shore    is    a    ship    whose   masts, 
55  feet  high,  are  just  disappearing  from  view? 


148  UNIFIED  MATHEMATICS 

9.    Using  the  figure  in  the  text,  find  an  approximation  for 
TA  in  miles  when  h  is  small  and  measured  in  feet. 


=v'- 


2h-r     f    h    \«_    /2  -  h  •  60     /    h   V 
5280      \5280y  ~"V      80          \52Wj 


16,     ,    /     h     V 

\/-/t4-[  -  ]  = 
*4        V5280; 


AT 


for    values   of    h    less    than   5  x  5280,   the  (  —  —  ]     can    be 
neglected. 

10.  Find  the  "  dip  "  of  the  horizon  and  the  distance  from 
the  balloon  for  h  =  100,  500,  and  1000  feet. 

11.  Find  the  distance  from  the  point  below  the  balloon  on 
the  earth's  surface  to  the  points  on   the   horizon  viewed   by 
the  observer  in  the  balloon. 

12.  According  to   the  approximate   formula   of   Huyghens 
the  length  of  a  circular  arc,  a,  is  connected  with  the  chord,  c, 
of  the  arc  and  the  chord,  h,  of  half  the  arc,  by  the  formula 

a  =  —      —  .     Compute  the  actual  length. 


CHAPTER   IX 

APPLICATIONS   OF  TRIGONOMETRIC   FUNCTIONS 

1.   Parallel  and  perpendicular  lines.  —  The  slope  of  the  line 
joining  fa,  yd  to  (x2,  y.2~), 


x.2  —  xl 

evidently  represents  the  tangent  of  the  angle  which  the  line 
joining  these  two  points  makes  with  the  positive  ray  of  the 
»-axis,  i.e.  the  angle  from  the  oj-axis  to  this  line.  We  have 
taken  P^fa,  y2)  to  the 
right  of  Pfa,  y^,  but 
obviously  interchanging 
Pzfa,  2/2)  and  Prfa,  yd 
simply  changes  sign  of 
both  numerator  and  de- 
nominator of  the  fraction 
representing  the  slope  m ; 
Q  is  in  each  figure  the 
point  fa,  yd,  and  P^Q 
and  QP2  have  like  signs 
if  PiP2  or  P^Pi  makes  a 
positive  acute  angle  with 
the  positive  ray  of  the 
or-axis  ;  and  P}  Q  and  QP2 
have  unlike  signs  in  the 
contrary  case  when  P^PZ 
or  PzPi  makes  a  negative  acute  angle  with  OX.  It  is  to  be 
noted  that  shifting  the  y-axis,  parallel  to  itself,  either  to  the 
right  or  to  the  left  does  not  affect  the  value  of  x^  —  x1}  since 

149 


150  UNIFIED  MATHEMATICS 

whatever  the  position  of  0,  A^A^  =  P1Q  =  OA2  —  OAl  =  x2  —  x\  ; 
similarly  no  change  is  made  in  the  value  of  the  slope  by 
shifting  the  <»-axis  parallel  to  itself,  up  or  down. 

Given  y  =  m^  +  k,  any  straight  line,  raj  represents  the 
tangent  of  the  angle  which  this  line  makes  with  the  positive 
ray  of  the  #-axis.  Any  parallel  line  has  the  same  slope  ; 
ra2  =  ?%  for  two  parallel  lines.  Any  perpendicular  line  has 
the  slope  angle 

««,  =  90°  -f  «!  ;  tan  <%  =  tan  (90°  +  ttl)=  —  cot  ^  = 


tan  ax 

whence  m2  =  —  —  .     Of  two  parallel  lines  the  slopes  are  equal, 

mx 

and  of  two  perpendicular   lines  the  slope  of   the  one  is   the 
negative  reciprocal  of  the  slope  of  the  other,  i.e. 

ra2  =  --  or,  by  solving,  ml  =  --  • 
ml  m2 

Given  y  =  mx  +  b,  any  family  of  parallel  lines  of  slope  m. 

y  =  —  —  x  +  k  represents  the  family  of  perpendicular  lines. 
m. 


Illustrative  problem.  —  Given  _3x  +  4y  —  1  =  0,  find  the  slope,  the 
parallel  line  through  the  origin,  the  family  of  perpendicular  lines,  and 
the  perpendicular  line  through  (—1,  5). 


=—  f,  0=-36°  52'. 
y  =  —  f  x  is  the  parallel  line  through  the  origin. 
Derive  this  both  from  y  =  mx  +  b  and  y  —  yl  =  m(x  —  x\). 

1  4 

The  perpendicular  line  has  the  slope,  m2=  --  -  =  +  -  • 

MI 

y  =  |  x  +  k  is  the  family  of  perpendicular  lines. 
y  —  5  =  |(x  +  1)  is  the  perpendicular  line  through  (—  1,  5). 

EXERCISES 

1.   Write  the  equations  of  the  sides  of  the  triangles  used  in 
finding  the  functions  of  30°,  45°,  and  60°. 


APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS   151 


Acceleration  down  a  plane, 
g  sin  «. 


2.  Gravity  imparts  to  a  falling  body  a  vertical  velocity  of 
32 1  feet  per  second,  with  t  seconds  as  time  during  which  the 
body  has  fallen ;  on  a  smooth  in- 
clined plane  gravity  imparts  a  ve- 
locity of  32  t  •  sin  a  where  a  is  the 

angle  of  inclination  of  the  plane. 
Find  the  velocity  imparted  at  the 
end  of  1  second  to  a  body  sliding 
(without  friction,  assumed)  on  an 
inclined  plane  of  slope  10°,  20°, 
30°,  40°,  50°,  ...  to  90°. 

3.  In    a    freely    falling    body 

s  =  16  tz ;    while  on  a  plane  s  =  16 12  •  sin  a ;  find  s  for  t  =  10, 
a  =  30°,  45°,  and  60°. 

4.  To   pull   the   body   up   the   plane   requires    a   force   of 
TFsin  a  +  kW-  cos  a,  where  A;  is  a  constant  dependent  upon 
the  friction.     Find  the  force  to  pull  a  weight  of  1000  Ib.  up 
an  incline  of  30°,  Jc  =  ^. 

5.  Find  the   slope  of  the  line   joining  (—3,  7)  to  (5,  9); 
find  the  middle  point  of  this  line  ;  find  the  equation  of  the 
perpendicular  bisector  of  the  segment. 

6.  Write  the  equation  of  the  line  through  (—3,  5)  making 
an  angle   tan"1  T5^-   (m  =  ^2-)  with   the  a-axis,    and  write  the 
equation  of  the  perpendicular  from  (1,  8)  to  this  line. 

7.  Find  the  foot  of  the  perpendicular  line  found  in  prob- 
lem 6  and  then  find  the  distance  between  (1,  8)  and  the  origi- 
nal line,  using  the  distance  formula. 

8.  Find  the  slope  angles  in  degrees  and  minutes  of   the 

following  lines : 

(a)  5  y  —  12  x  —  7  =  0, 

(6)   127/  +  5x-3  =  0, 

(c)  x  -  y  -  5  =  0, 

(d)  3x-y-  8.=  0. 

9.  Find  lines  through  (1,  5)  parallel  and  perpendicular  to 
each  of  the  lines  in  the  preceding  exercise. 


152 


UNIFIED  MATHEMATICS 


-O 


10.  Find  the  perpendicular  bisectors  of  the  sides  of  the 
triangle  formed  by  the  three  lines  given  by  the  equations, 
5y— 12  #— 7=0,  12y+5o;— 3=0,  and  x+y— 5=0.  Find  the 
area  of  this  triangle  graphically  and  analytically. 

2.  Projections  of  vectors.  —  OP  has  been  designated  by  r,  for 
radius  vector  of  the  point  P.  The  line  OP  has  magnitude,  given 
by  r,  and  direction,  given  by  the  angle  6.  We  may  use  this 

system  of  representation  to  repre- 
sent velocities,  forces,  and  other 
physical  quantities.  As  a  velocity 
this  vector  may  be  resolved  into 
two  component  velocities,  repre- 
sented by  OA  and  OB.  OA  repre- 
sents the  velocity  in  the  x  direction, 
x  =  r  cos  6 ;  OB  represents  the  y 
velocity,  r  sin  6,  the  vertical  com- 
ponent of  the  velocity  of  a  body 
moving  with  velocity  represented  by 
OP.  The  projection  of  any  vector 
upon  a  directed  line  is  defined  as 
the  directed  distance  between  the  perpendiculars  dropt  from 
the  extremities  of  the  given  vector  upon  the  line ;  it  is  given 
by  v  cos «  wherein  v  represents  the  vector  and  a  is  the 
angle  between  the  positive  rays  of  the  two  lines.  Since 
cos  (—  «)=  cos  a,  we  do 
not  need  to  distinguish 
between  the  two  lines, 
i.e.  the  angle  can  be 
taken  as  obtained  by  ro- 
tation from  the  given 
line  to  the  given  vector, 
or  vice  versa. 

It  is  a  fundamental  as- 
sumption that  any  two  vector  quantities  which-  may  be  repre- 
sented acting  together  at  the  same  point  may  be  replaced  by 


Components  of  a  vector 

OA  represents  the  x  com- 
ponent of  OP. 

OB  represents  the  y  com- 
ponent of  OP. 

OP  is  the  resultant  of  OA 
and  OB. 


Vector  parallelogram 


APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS   153 


a  single  vector  which  is  the  diagonal  of  the  parallelogram 
formed  by  the  two  given  vectors.  The  process  is  called 
vector  addition.  .  This  assumes  that  in  space,  for  example,  an 
imparted  velocity  S.  E.  of  50  miles  per  hour  increased  by  a 
velocity  N.  E.  of  30  miles  per  hour  produces  the  same  displace- 
ment whether  the  two  forces  which  produce  the  velocities  act 
together  for  one  hour,  or  whether  both  act  in  succession  each 
for  an  hour. 

The   projection   of  a  broken  directed   line   upon   a  given 
directed  line  is  the  same  as  the  projection  of  the  straight  line 
joining  the  ends  of  the 
broken  line. 

This  follows  from  the 
fact  that  on  a  directed 
line 


1 


whatever  the  relative  po- 
sitions of  MI,  M2,  and 
M3.  The  directed  length 

is  the  projection  of 

M»MZ  is  the  pro- 
jection of  P2P3,  MiMz  is 
the  projection  of  PiP3. 
The  physical  interpreta- 
tion is  simply  that  the 
total  component  in  the 

x  direction  (or  any  other)  imparted  by  two  (or  more)  vectors 
is  the  algebraic  sum  of  the  two  (or  more)  x  components  of 
these  vectors,  taken  separately. 

When  the  velocity  is  given  as  v,  vz  and  vy  are  commonly 
used  to  designate  the  x  and  y  components  of  the  velocity; 
evidently,  also 


Projection  of  a  broken  line  on  a  directed 
line 


vx  =  v  cos  6, 
v,,  =  v  sin  0. 


154  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  A  bullet,  muzzle  velocity  of  3000  feet  per  second,  leaves 
the  gun  elevated  at  an  angle  of  10°.     The  position,  neglecting 
air  resistance,  is  determined  at  the  end  of  t  seconds  by  the  two 
equations :  y  =  3m  t  gin  1Qo  _  16  ^ 

x  =  3000 1  cos  10°. 

Find  t  when  y  =  0  ;  when  y  =  5  ;  explain  the  two  values  in 
each  case.     Find  x  for  both  values  of  t  which  make  y  =.  0. 

2.  The   velocity    is    a    vector    resolved    into    components 
vx  =  v  cos  «  and  vy  =  v  sin  a.     Find  vx  and  vv  when  «  =  10°, 

20°,  30°,  45°,  60°. 

i 

3.  A  ship  sails  S.  E.  for  2  hours  at  8  miles  per  hour  and 
E.  N.  E.   (22f  off  East)  for  2  hours  at  6  miles  per  hour. 
Find  the  x  and  y  of  the  resultant  position. 

4.  The  propeller  imparts  to  a  steamer  a  velocity  of  8  miles 
per  hour  S.  E.  (—  45°)  and  the  wind  imparts  a  velocity  of 
E.  N.  E.  (+  221°)  of  6  miles  per  hour.     Find  the  position  at 
the  end  of  1  hour. 

5.  A  boy  runs  east  on  the  deck  of  a  steamer  at  the  rate  of 
20  feet  per  second  ;  the  steamer  moves  south  at  the  rate  of  15 
miles  per  hour.     Find  the  actual  direction  in  which  the  boy  is 
moving  and  his  total  velocity. 

6.  Find  the  velocity  in  miles  per  hour  of  a  point  on  the 
earth's  surface  due  to  the  rotation  of  the  earth  on  its  axis ; 
find  the  velocity  per  second  due  to  the  revolution  about  the 
sun ;  compare,  and  note  that  the  resultant  can  never  be  greater 
than  the  sum  nor  less  than  the  difference  of  the  two.     Take 
values  only  to  3  significant  figures ;  3960  mi.  =  r ;  93,000,000 
miles  as  distance  from  sun. 

7.  The  United  States  rifle,  model  1917,  has  a  muzzle  velocity 
of  2700  feet  per  second.     Find  the  horizontal  velocity  of  the 
bullet  when  the  angle  of  elevation  is  1°,  10°,  20°,  30°,  and  45° 
respectively. 


APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS   155 

3.  Normal  form  of  a  linear  equation.  —  The  slope-intercept, 
point-slope,  and  two-point  formulas  correspond  to  the  fact 
that  a  straight  line  is  determined  when  one  point  on  the  line 
[(0,  A;)  or  (xl}  y^)  respectively]  and  the  direction  of  the  line  are 
given,  or  when  two  points  are  given.  A  straight  line  may  be 
determined  in  many  other  ways  ;  one  method  which  gives  a 
further  useful  form  of  the  equation  of  the  straight  line  deter- 
mines the  line  in  terms 
of  the  length  and  direc- 
tion of  the  perpendicular 
from  the  origin  upon  the 
line. 

Thus  if  a  perpendicu- 
lar from  the  origin  upon 
a  given  line  is  5  units 
long,  and  makes  an  angle 
of  120°  with  the  x-axis 
(positive  ray)  geometri- 
cally we  construct  the 
line  by  constructing  the 

ray  6f  120°  and  upon  it  taking  a  length  of  5  units.  At  the 
extremity  of  this  line  of  5  units  length  a  perpendicular  is 
drawn  which  is  the  required  line.  The  point  N  is  readily 

found  to  be  (5  cos  120°,  5  sin  120°)  and  the  slope  is 
therefore  the  equation  of  the  line  to  be  found  is 


A  line   determined  by  the   normal  to  it 

from  the  origin 
Normal  length,  5  ;  a  =  120°. 


tan  120°' 


y-  5  sin  120°  = 


-1 


tan  120' 


,  (a?  -5  cos  120°). 


tan  120°  =  Sm          ;  substituting,  clearing  of  fractions,  trans- 


posing 


cos  120° 


x  cos  120°  +  y  sin  120°  -  5  (sin*  120°+  cos2  120°)=  0, 
x  cos  120°  +  y  sin  120°  -5=0,  since  sin2  a  +  cos2  a  =  1, 


5  =  0. 


156 


UNIFIED  MATHEMATICS 


In  general,  given  the  normal  ON  to  the  line  from  the  origin, 
of  length  p,  and  making  angle  a  with  OX,  the  extremity  N  is 

(p  cos  a,  p   sin  a)  ;    the   slope   is 

— ,  and  the  equation  becomes 
sin  a 

x  cos  a.  +  y  sin  a  —  p  =  0, 

^>  is  taken  as  a  positive  quantity 
just  as  r  has  been  taken.  Evi- 
dently if  p  =  0, 

x  cos  «  +  y  sin  «  =  0 

represents  a  parallel  line  through 
the  origin.  Evidently  also  for 
parallel  lines  on  opposite  sides  of 

the  origin  the  angles  a  and  a'  differ  by  180°  ;  i.e.  a'  =  180°  +  a, 

whence  •      •  _  _    • 

cos  a'  =  —  cos  a. 


x  cos  «  +  y  sin  a  —p  = 
Normal  form. 


A 


The  projection  on  ON  } 
ofOM  +  MP          ] 


the  projection  on  ON 
ofOP 


4.   Normal  form  derived  by  projection.  —  We  have  shown  that 
the  projection  of  any  broken  line  upon  any  given  line  is  the 


APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS    157 

same  as  the  projection  upon  the  given  line  of  the  vector  join- 
ing the  ends  of  the  broken  line.  Let  P(x,  y)  be  any  point 
on  the  line  whose  equation  is  sought;  drop  PM  the  perpen- 
dicular from  P  to  the  avaxis;  the  projection  of  the  broken 
line  OM  +  MP  on  the  normal  OjVis  equal  to  the  projection  of 
OP  on  ON.  Now  OM  =  x  makes  the  angle  a,  by  hypothesis, 
with  ON,  and  MP  makes  the  angle  a  —  90°  ;  hence  the  projec- 
tion of  OM  on  ON  is  x  cos  a  (OA,  negative  in  the  figure 
since  a  is  obtuse)  and  of  MP  on  ON  (AN  in  the  figure)  is 
y  cos  (a— 90°) ;  the  projection  of  OP  on  ON  is  ON  itself,  or  p; 
further  y  cos  (a  —  90°)  =  y  cos  (90°  —  a)  =  y  sin  a.  Then,  since 
projecting  on  the  line  ON, 

projection  of  OM  +  projection  of  MP=  projection  of  OP, 
we  have 

x  cos  a  +  y  sin  a  =  p,  whence  x  cos  a  +  y  sin  a  —  p  =  0. 

5.    To  put  the  equation  of  a  straight  line  in  normal  form.  — 
Let  the  given  equation  be  3  a;  —  4  y  -f  7  =  0,  and  let 
x  cos  a  +  y  sin  a  —  p  =  0  be  the  same  equation  in  normal  form. 

If  these  two  equations  represent  the  same  line,  these  lines 
must  have  the  same  slope  and  the  same  y  (or  x)  intercept. 

—  cos  a  __  3     1  _     p 
sin «        44      sin  a 

cos  a  =  ^—  sin  a. 
4 

cos2  a  =  T9^  sin2  a. 

But  cos2  a  =  1  —  sin2  a, 

whence  1  —  sin2  a  =  T9g  sin2  a  ;  T|  sin2  a  =  1 ;  sin  a  =  ±  |. 

p  =  +  %  sin  a,  whence  since  p  is  to  be  positive,  sin  a  must  be 
taken  as  positive.  Hence  sin  a  =  +  £,  p  =  -£ ;  cos  a  =  —  | ; 
and  thus  the  normal  form  is  —  fx  +  fy  —  |  =  0.  This  equa- 
tion is  obtained  by  dividing  each  member  of  the  original 
equation  by  —  5. 


158 


UNIFIED  MATHEMATICS 


27) 


In  general  to  put  Ax  -f  By  +  C  =  0  in  the  normal  form, 
a;  cos  a -f-t/ sin  a  —  p  =  0,  one  must  multiply  through  by  some 
quantity  k,  so  that  fc^l  =  cos«,  A,-JB  =  sin«,  and  kC  =  —  p-, 
kC=—p  shows  that  k  must  be  chosen  opposite  in  sign  to  C; 
squaring  both  members  of  the  first  two  equations  and  adding 

gives  kz(A-  +  B2}  —  1,  whence  k  —  ±  —  =,  of   which  the 

sign  is  taken  as  opposite  to  C.  vA2  +  B2 

RULE.  —  To  put  an  equation  Ax  +  By  +  C=  0  in  normal  form 
divide  through  by  —  ±  \f  A*  •+  B2,  with  the  sign  taken  opposite  to 
that  of  the  constant  term. 

6.  To  find  the  perpendicular  distance  from  a  point  to  a  line.  — 
In  solving  this  problem  one  considers  the  various  forms  of  the 

straight  line 
which  may  be 
employed.  Evi- 
dently the  normal 
form  is  most 
hopeful  for  use, 
since  it  involves 
the  perpendicular 
distance  of  the 
given  line  from 
the  origin. 
Through  the 
point  PI(XI}  yO 
draw  a  line  paral- 
lel to  the  given 
line ;  evidently 
the  difference  be- 
tween the  nor- 
mals to  the  two 
lines  gives  the 

distance.      Three  possibilities  must  be  considered :  1.   P1}  on 
the  opposite  side  of   the  given  line  from  the  origin;   2.    Plt 


'0* 


Distance  of  a  point  from  a  line 


APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS   159 

on  the  same  side  of  line  as  0,  the  origin,  but  such  that  the 
normal  angle  is  the  same,  i.e.  so  that  the  parallel  line  through 
PI(XI,  yt)  falls  on  the  same  side  of  0  as  the  given  line,  P\  011 
the  figure  ;  3.  P1}  on  the  same  side  as  the  origin,  but  the 
normal  angle  increased  (or  diminished)  by  180°,  designated  by 
P"i  on  the  figure. 

Let  x  cos  a  +  y  sin  a  —  p  =  0  be  the  equation  of  the  line. 

1.  x  cos  a  -f-  y  sin  a  —  (x^  cos  a  -|-  y±  sin  a)  =  0  is  the  parallel 
line  through  PI(X},  y^),  since  this  equation  is  evidently  in 
normal  form  and  the  line  passes  through  (xly  y^). 

ONZ  =  Xi  cos  a  +  y!  sin  «. 
d  =  ON%  —  ON  =  xt  cos  a  -\-yi  sin  «  —  p. 


The  perpendicular  distance  is  obtained  then  by  writing  the 
equation  in  normal  form  and  substituting  for  (a/!,  y\)  the 
coordinates  of  the  given  point.  Evidently  if  PI  (xl}  y^)  is  on 
the  line,  this  gives  also  the  correct  distance,  which  is  then  zero. 

2.  x  cos  a+y  sin  a  —  (x}  cos  a-\-yl  sin  a)  is  the  equation  of  the 
parallel  line  ;  again,  ON'2  =  x±  cos  a  -f-  y±  sin  a. 

d=ON-  ON'  2  :  whence    —d=  ON'2  -  ON 

=  Xi  cos  a  +  yi  sin  a—  p. 

The  same  rule  holds,  but  the  distance  in  this  case  is  negative. 
Evidently  the  rule  holds  if  ON1^  is  0. 

3.  «'  =  180  -f  a  ;  cos  «'  =  —  cos  a,  sin  a'=  —  siu  a. 

To  write  the  equation  of  the  parallel  line  in  normal  form,  the 
coefficients  of  x  and  y  must  both  be  the  negatives  of  the  coeffi- 
cients of  x  and  y  in  the  given  equation. 

x(—  cos  a)  +  y(—  sin  «)  —  (—  a^cos  a  —  y^  sin«)=0  is  the 
equation  of  the  parallel  line  in  normal  form. 


=  —  (x{  cos  a  +  £/!  sin  a). 
d  =  ON"2  +  ON=  —  a?!  cos  «  —  y^  sin  a  +p, 
or  —  d  =  #1  cos  a  +  yt  sin  a  —  p. 


160  UNIFIED  MATHEMATICS 

RULE.  —  To  obtain  tin-  iHxtmice  from  a  point  to  a  ////<• 
the.  equation  in  normal  form,  substituting  therein  for  #  and  y  tlie 
coordinates  of  the  given  point.  The  resulting  number  gives  the 
distance  as  positive  if  the  point  and  the  origin  lie  upon  opposite 
sides  of  the  given  line,  as  negative  if  Pl  and  0  are  upon  the  same 
side  of  the  given  line. 

x  cos  «  +  y  sin  «  —  p  represents  the  perpendicular  distance 
then  from  P(x,  y)  to  the  line  x  cos  «  +  y  sin  «  —  p  —  0.  For 
all  points  on  one  side  of  this  line  the  expression  is  positive, 
and  on  the  other  side,  crossing  the  line  to  the  origin  side,  the 
expression  is  negative. 

A  line  which  passes  through  the  origin,  p  =  0,  will  be  said 
to  have  its  equation  in  normal  form  when  sin  a  is  taken  as 
positive,  i.e.  when  the  coefficient  of  y  is  made  positive.  Points 
on  this  line  make  x  cos  a  +  y  sin  a  =  0  ;  points  above  the  line 
make  the  expression  x  cos  a  -f  y  sin  a  positive,  and  points  be- 
low the  line  make  it  negative. 

Thus  3x—  4  y  =  0  is  written  ^—  x  -f  -  y  =  0,  or 

5  o 

-3a  +  47/_ 


to  be  in  normal  form.  The  perpendicular  distance  from  any 
point  to  such  a  line  will  be  positive  for  points  above  the 
line,  and  negative  for  points  below  the  line. 

7.   Bisector  of  the  angle  between  two  lines.  —  Geometrically 
the  bisector  of  an  angle  is  the  locus  of  the  points  equidistant 
from  the  two  sides  of  the  angle  ;  analytically  we  express  the 
condition  that  two  distances  should  be  equal  to  each  other. 
Let  the  equations  be  given  in  normal  form,  as 
x  cos  «t  +  y  sin  «!  —  pl  =  0  and  x  cos  a^  +  y  sin  a,  —  Pz  =  0. 

Let  P(x,  y)  represent  any  point  on  either  bisector  of  the 
given  angle  ;  analytically 

x  cos  «!  +  y  sin  a^  —  pL  =  ±  (x  cos  a,  +  y  sin  a«  —  p»~). 


APPLICATIONS   OF  TRIGONOMETRIC   FUNCTIONS    161 


Bisector  A,  in  the  opening  which  includes  the  origin,  is  ob- 
tained by  taking  the  +  sign  since  both  perpendiculars  are  of 
the  same  sign  for  points  on  A.  Bisector  B  is  obtained  by 
taking  the  negative  sign  since 
any  point  on  B  is  on  the 
same  side  as  the  origin  with 
respect  to  one  of  the  lines, 
and  on  the  opposite  side  with 
respect  to  the  other ;  hence 
if  PMl  comes  out  negative, 
PM2  will  be  positive  (by  the 
formula)  and  the  equality 
will  be  obtained  by  putting 
PMi  =  -  PM2. 

Just  as  the  two  axes  divide 
the  plane  into  4  quadrants  in      Bisectors  of  ^  angles  between  two 
which  the  distances   to   these  lines  given  normal  form 

axes  are  +  +,   —  +,   —  — , 

and  +  —  respectively,  so  any  two  lines  in  the  plane  divide 
the  plane  into  4  sections  in  which  the  perpendicular  distances, 
as  given  by  our  formula,  to  these  lines  are  ++,+  —,-  —  > 
and  —  +•  respectively.  The  +  +  and  —  —  sections  are 
separated  by  the  -f  —  and  —  +  sections  respectively,  as  it  is 
evident  that  you  pass  from  +  +  to  +  —  by  crossing  the 
second  line. 

The  bisector  of  the  -f-  +  and  —  —  opening  is  given  by 
equating  the  left-hand  members  of  the  equations  of  the  two 
lines  in  normal  form;  the  bisector  of  the  -f-  — ,  -  +  opening 
is  obtained  by  equating  the  one  to  the  negative  of  the  other 
left-hand  member. 

If  one  of  the  lines  passes  through  the  origin,  or  if  both 
do,  then  the  above-mentioned  convention  is  necessary  to  es- 
tablish the  part  of  the  plane  in  which  the  left-hand  member  of 
the  equation  of  the  line  is  positive.  It  is  customary  to  make 
sin  a  positive,  which  makes  the  portion  of  the  plane  above  the 
line  the  positive  side,  i.e.  the  coordinates  of  any  point  above 


162  UNIFIED  MATHEMATICS 

the  line  when  substituted  in  the  given  equation  give  a  positive 
value,  and  of  any  point  below  the  line  give  a  negative  value. 

PROBLEMS 

1.  If  a  line  makes  an  angle  of  30°  with  the  ar-axis  what 
angle  does  the  normal  to  the  line  make  with  the  x-axis  ? 

2.  What  is  the  slope  of  the  line,  y  =  2  x  -+•  5  ?     What  is  the 
slope  angle  ?     What  is  the  slope  of  the  normal  to  this  line  ? 
AVhat  is  the  angle  which  this  normal  makes  with  the  z-axis  ? 
Find  from  the  tangent  of  the  angle  made  by  the  normal  with 
the  x-axis  the  sine  and  cosine  of  the  same  angle.     Write  the 
equation  in  normal  form  and  interpret  the  constants. 

3.  Given  a  and  p,  as  below,  slope  angle  of  the  normal  and 
length  of   the  normal    from  the  origin   to  the   line,  find  the 
equations  of  the  lines,  and  draw  the  lines : 

a.  a  =      30°,        p  =  5. 

b.  «  =  -30°,        p  =  5. 

c.  a=    150°,        p  =  5. 

d.  a=    210°,        p  =  5. 

e.  a=    137°,        p  =  5. 
/.    «=    137°,        p  =  W. 
g.    a  =  -  63°,        p  =  10. 
k.    a  =    223°  15',  p  =  8. 

4.  If  a  remains  equal  to  40°  and  p  varies,  what  series  of 
lines  will  be  obtained  ?  if  p  remains  equal  to  5,  and  a  varies, 
what  series  of  lines  will  be  obtained  ? 

5.  Write  the  following  equations  in  normal  form : 

a.   3x  —  4y—  5  =  0. 
6.   5x  - 


c.  x 

d.  3z-5#-4  = 

e.  y  =  2  x  —  14. 


35 


APPLICATIONS  OF  TRIGONOMETRIC   FUNCTIONS    163 

6.  Find  the  distances  of  the  points  (1,  5),  (2,  3),  (0,  5), 
(0,  —  5),  (—2,  —3),  and  (—3,  7)  from  each  of  the  lines  in 
the  preceding  problem. 

7.  What  is  the  distance  of  the  point  (x,  y)  from  the  line 
3x  —  4y  —  5  =  0?   Under  what  circumstances  does  the  formula 
give  a  negative  value  for  this  distance  ?     What  is  the  distance 
of  any  point    (x,  y)   from   5z-fl2?/-f8  =  0?     What    does 
equating  these  two  expressions,  i.e.  the  left-hand  members  of 
each  normal  form,  give  ?     Interpret  on  the  diagram.     What  is 
obtained  by  setting  one  of  these  expressions   equal    to   the 
negative  of  the  other? 

8.  Find  the  bisectors  of  the  angles  between  the  •  following 
pairs  of  lines : 

a.  3x  +  4y—  5=0  and  12  x  —  5  y  —  10  =  0. 

b.  y—2x—5=0  and  2x  +  y  +  7  =  0. 

c.  y  —  2x— 5  =  0  and  3y+x  —  8=0. 

d.  y  -2x  =  Osmd3y  +  x-8  =  Q. 

e.  y  —  2  x  =  0  and  3  y  -f-  x  =  0. 

9.  Find  the  distance  of  the  points  (1,  —  3),  (3,  0),  (3,  -  7), 
and  (0,  —  8)  from  each  of  the  lines  in  the  preceding  problem. 

10.   Find  the  distance  between  the  following  pairs  of  paral- 
lel lines  : 

a.  y  =  2  x  —  7, 
y=2x  +  3. 

b.  4?/-3a  =  5, 
4y-3a-16  =  0. 

c.  4  y  —  3  x  =  0, 
4y-3z  —  16  =  0. 

d.  x  +  2y-7  =  0, 
2a  +  4y  +  17  =  0. 

e.  7.2 x  +  8.3  y-  15  =  0, 

-    8  =  0. 


164  UNIFIED  MATHEMATICS 

11.  In  problem  8  show  that  each  bisector  obtained  is  one 
of  the  pencil  of  lines  through  the  point  of  intersection  of  the 
given  two  lines. 

12.  Find  the  area  of  the  triangle   having  as  vertices  the 
following  points : 

a.  (3,  4),  (0,  0),  and  (0,  8). 

b.  (3,  4),  (0,  0),  and  (10,  2). 

c.  (1,  1),  (4,  5),  and  (7,  -3). 

13.  Find  the  area  of  the  triangle  formed  by  the  three  lines : 

3x  +  ±y-    5  =  0, 

12  x  -  5  y  -  10  =  0, 

and       4x  —  3y  —    7  =  0. 

14.  What  is  the  distance  of  any  point  (x,  y)  from  the  point 
(0,  0)  ?     What  is  the  distance  of  any  point  (x,  y)  from  the 
line  x  —  5  =  0  ?     Equate   these  two  expressions  for  distance 
and  simplify.     The  resulting  equation   has  for  its  graph  all 
points  which  are  equally  distant  from  the  point  (0,  0)  and  the 
line  x  —  5  =  0. 

15.  Find  the  locus  of  all  points  which  are  equidistant  from 
the  point  (0,  0)  and  the  line  y  —  8  =  0.     Let  (x,  y)  represent 
any  point  satisfying  the  given  condition. 

16.  Find  the  locus  of  all  points  at  a  distance  10  from  the 
point  (0,  0) ;  from  (1,  —  3).     Find  the  locus  of  all  points  at  a 
distance  10  from  the  line  3x—  4  y  —  7  =  0;  at  a  distance  —  10 ; 
explain  graphically. 

17.  Find   the    locus   of    all    points   equally   distant    from 
3  x  —  ±y  — 5=0  and  from  (1,  —  5). 

18.  In  problem  12  find  the  equations  of  the  three  bisectors 
of  the  angles  of  the  triangle  formed ;  find  the  perpendiculars 
from  the  vertices  to  the  opposite  sides ;  find  the  perpendicu- 
lar  bisectors   of   the   sides ;  show  that  in  each  instance  you 
have  three  lines  which  have  a  point  in  common. 


APPLICATIONS  OF  TRIGONOMETRIC  FUNCTIONS   165 

19.  What  points,  when  the  coordinates  are  substituted  for 
x  and  y,  make  the  expression  4  y  —  3  x  —  5  positive  ?     What 
points   make  this  expression  zero?     What  points  make   this 
expression  negative  ?     Locate  three  points  of  each  type,  plot 
and  discuss. 

20.  Substitute  in  the  expression   x2  +  y1  —  25   for  x  and  y 
the  coordinates  of  the  points  (0,  3),   (±  3,  2),  (±  1,  4).     Plot 
these  points.     Substitute  (0,  ±  5),  (±  3,  ±  4),  (±  4,  ±  3),  and 
(±  5,  0).     Plot.     Substitute  also  (0,  8),  (±  7,  0),  (5,  3),  and 
(±4,  6).     Note  that  the  graph  of  x2  +  y2  —  25  =  0  separates 
the  plane  into  two  parts ;  in  the  one  part  inside  this  curve  are 
all  points  whose  coordinates  substituted  for  x  and  y,  respec- 
tively, make  the  expression  x2  +  yz  —  25  negative,  and  in  the 
part  outside  lie  all  points  which  make  this  expression  positive. 


CHAPTER   X 

ARITHMETICAL   SERIES   AND   ARITHMETICAL 
INTERPOLATION 

1.  Definition  of  an  arithmetical  series.  —  In  the  table  of  natu- 
ral sines  the  values  of  the  sines  of  21°  to  22°  are  given  as 
follows, 

sin  21°         siu  21°  107         sin  21°  2<K         sin  21°  30'         sin  21"  4<X         sin  21°  50'         sin  22" 
.8684  .3611  .3638  .3665  .3692  .8719  .3746 

It  is  to  be  noted  that  each  value  differs  from  the  preceding  by 
.0027,  and  each  angle  differs  from  the  preceding  by  10'- 
Either  of  these  sequences  of  numbers  with  a  constant  differ- 
ence between  each  number  and  the  preceding  is  termed  an 
arithmetical  series  ;  the  continuation  of  the  lower  series  by  the 
successive  addition  of  .0027  gives  indefinitely  further  values 
of  the  arithmetical  series,  but  gives  only  5  following  sines. 
Of  the  series  of  numbers  given  to  four  decimal  places  the 
values  of  the  sines  of  angles  which  increase  by  intervals  of  10' 
it  happens,  for  reasons  which  will  be  further  discussed  below, 
that  twelve  values  beginning  with  the  sine  of  21°  coincide  with 
the  first  twelve  terms  of  an  arithmetical  series  ;  the  sine 
series  must  not  be  confused  with  the  arithmetical  series,  as  it 
is  only  arithmetical  in  limited  intervals  and  then  only  when 
approximate  values  are  used.  Thus  if  five  place  values  of  the 
sines  of  the  angles  above  were  given  the  series  would  no 
longer  be  arithmetical. 

The  type  form  of  arithmetical  series  is 


a        a  +  d        a  +  Zd        ii  +  Sd      ....      a  +  9d      ....      a+(n  -  2)d        a+(n  -  l)d 
1st  term     2d  3d  4th         ....         10th         ....     (n-l)thterm         «th  term 

each    term   is   d    greater   than   the   preceding   term    of    the 
series. 

166 


ARITHMETICAL  SERIES  AND  INTERPOLATION     167 

ln  =  a  -f  (n  —  l)rf  ;  by  ln  we  designate  the  nth  term  of  such 
a  series.  It  is  evident  from  the  definition  that  the  tenth 
term  in  such  a  series  is  a  -(-  9  d,  since  the  common  difference 
d  appeared  first  in  the  second  term  and  one  further  d  was 
added  in  each  subsequent  term. 

2.  Last  or  nth  term,  and  sum.  —  Strictly  we  should  prove  by 
a  process  called  mathematical  induction,  that  the  formula, 

lu  =  a  +  (H  —  T)d, 

always  represents  the  »zlh  term.  Evidently  for  n  =  1  this 
does  represent  our  first  term  ;  for  n  =  2  the  expression  does 
represent  our  second  term  ;  for  n  =  3  the  expression  a  +  2  d 
does  represent  our  third  term  ;  let  us  suppose  that  for  71  this 
does  represent  our  «th  term,  then  our  (n  +  l)th  term,  which  is 
d  greater,  must  be  ln  +  d  =  a  -f  (n  —  l)c?  +  d  =  a  +  nd;  now 


the  formula  gives  ln+l  =  a  +  (n  +  1  —  l)rf  =  a  .  +  nd  ;  hence 
if  this  formula  is  correct  for  the  nth  term,  the  formula  is  cor- 
rect for  the  next,  the  (n  +  l)th  term.  However,  we  know  that 
the  formula  is  correct  for  the  third  term,  hence  it  is,  by  our 
theorem  just  stated,  true  for  the  next,  the  fourth  term  ;  since 
it  is  true  for  the  fourth  it  is,  by  the  theorem,  true  for  the  fifth  ; 
so  for  every  subsequent  term. 

Frequently  the  sum  to  n  terms,  sn,  of  such  a  series  is  desired. 
To  obtain  a  simple  expression  for  sn,  we  proceed  as  follows  : 

sH  =  a+  (a  +  d)  +  (a  +  2d)+  ...a  +  9d+  •••  a  +  (n  —  l)d 
or  ln  ;  reversing  the  series  gives, 

sn  =  ln+  (ln-d)  +  (ln-2d)  +  ...Jn_9d+  ...Zn-(n-l)d; 
adding, 

)  +  -(a  f  /„)  +  •••(a  +  Zn); 


Fundamental  formulas  ; 


168  UNIFIED  MATHEMATICS 

3.  Practical  importance.  —  Arithmetical  series  are  of  great 
importance  because  of  their  occurrence  in  practical  problems, 
and  because  they  are  fundamental  in  the  applications  of 
mathematics  to  statistical  problems.  In  physical  problems 
involving  time,  the  time  is  commonly  measured  at  the  end  of 
equal  intervals,  giving  an  arithmetical  series  for  the  time ;  in 
the  tables  of  logarithms  our  numbers  increase  arithmetically, 
and  so  in  the  tables  of  trigonometric  functions  the  angles 
increase  arithmetically.  Refinement  of  measurement  is  com- 
monly made  by  subdividing  the  unit  of  measurement  into 
smaller  equal  intervals,  giving  new  arithmetical  series. 

PROBLEMS 

1.  Find  the  tenth  and  the  twentieth  terms  of  the   series, 
1,  3,  5,  7,  -  ;  find  the  (n  -|-  l)th  term. 

2.  Find   the   sum   to   10   and   to   20   terms   of   the   series 
1,3,5,-.. 

3.  Show  by  mathematical  induction  that  the  sum  of  the  first 
n  odd  numbers  is  ?i2,  by  showing  that  if  the  sum  is  n-  then  the 
sum  of  the  first  (n  -+-  1)  odd  numbers  is  (n  -f  I)2. 

4.  Solve  I  =  a  +  (n  —  l)cZ,  for  d  ;  solve  for  a ;  solve  for  d. 

5.  Given  I  =  235,  d  =  7,n  =  40,  find  a. 

6.  Given  I  =  235,  d  =  7,  a  =  5,  find  d. 

7.  Given  d  =  7,  a  =  5,  n  =  40,  find  I. 

8.  One  hundred  men  increase  uniformly  in  height  from  5.01 
feet  to  6  feet  by  .01  of  a  foot,  find  the  total  height ;  if  their 
weights  increase  uniformly  by  half-pounds  from  110  pounds, 
find  the  total  weight  of  the  group,  and  the  average  weight. 

9.  On  an  inclined  plane,  angle  of  30°,  a  ball  rolls  approxi- 
mately 8  feet  in  1  second,  24  feet  in  the  second  second,  40  feet 
in  the  next,  and  in  every  second  16  feet  more  than  in  the  pre- 
ceding second.     Find  the  distance  a  ball  travels  in  5  seconds  ; 
in  10  seconds.     This  formulation  neglects  the  energy-loss  due 
to  rolling. 


ARITHMETICAL  SERIES  AND  INTERPOLATION     169 


10.  On  a  hill  inclined  at  30°  a  bob-sled  moves  approximately 
according  to  the  law  of  the  rolling  ball  in  the  preceding 
problem.  Find  the  length  of  time  to  cover  1000  feet.  Find 
the  distance  covered  in  the  last  second  of  the  slide.  Find 
the  average  velocity  during  the  slide,  and  the  average  velocity 
during  the  last  second.  Reduce  velocity  to  miles  per  hour. 

NOTE.  —  Average  velocity  is  simply  the  space  covered  divided  by  the 
time  required. 

4.  Graphical  representation.  —  The  arithmetical  series  is 
represented  graphically  by  the  straight  line,  and  conversely 
any  straight  line  represents  an  arith- 
metical series.  For  this  reason  the 
interpolation  processes  explained 
above  under  logarithms  and  under 
trigonometric  functions  are  some- 
times termed  "  straight-line  interpola- 
tions "  ;  the  process  is  correct  in  those 
small  intervals  in  which  the  curve 
representing  the  function  is  approxi- 
mately a  straight  line. 

For  the  integral  values  of  x,  from 
0,  1,  2,  3,  •••  up  to  n  —  1,  the  ordi- 
nates of  the  line 

y  =  dx  +  a 

represent  graphically  the  terms  of 
the  type  arithmetical  series,  a,  a  +  d, 
a  +  2  d,  •  • ;  for  values  of  x  from  0, 
n-1 


up  to 


10 


the  or- 


/ 

i 

Y 

/ 

d 

\> 

/ 

U 

/ 

d 

<s 

t 

-A 

>Sv 

a 

*/ 

•"/ 

d 

>a 

Id 

i 

I 
-^ 

/ 

d 

» 

Al 

\ 

a\ 

1 

! 

0 

is 

34 

5  < 

17* 

\  X 

dinates   represent   the   terms   of  an 
arithmetical  series  with  first  term  a 


and  the  common  difference  — . 


To 


The  ordinates  of  y  =  dx  +  a 
at  x  =  0,  1,  2,  3,  •••  repre- 
sent terms  of  an  arithmet- 
ical series 


any  series  of  equal  increases  or  increments  given  to  x  there 
correspond  a  series  of  equal  increments  given  to  the  ordinates ; 


170 


UNIFIED  MATHEMATICS 


this  depends  upon  the  theorem  of  plane  geometry  that  if  a 
series  of  parallel  lines  cut  off  equal  parts  on  one  transversal 

they  do  on  every  transversal,  and 
this  theorem  is  equally  true  for 
any  straight  line  in  the  plane. 

5.  Interpolations  in  sines  and 
other  functions.  —  The  value  of  the 
sines  of  the  angles  are  given  by 
the  corresponding  ordinates  in  a 
circle  of  radius  unity,  or  the  or- 
dinates divided  by  100  in  a  circle 
of  radius  100,  or  the  ordinates  di- 
vided by  1000  in  a  circle  of  radius 
1000. 

On  our  diagram,  with  radius 
100  the  straight  line  joining  the 
end  of  the  ordinate  corresponding 
to  20°  to  the  end  of  the  ordinate 
at  30°  does  not  differ  materially 
from  the  circular  arc  connecting 
these  points.  Were  we  to  plot 
these  angles  in  a  circle  of  radius 
1000  the  points  of  intersection 
would  appear  as  in  the  second  part 
of  the  diagram,  lettered  AB,  and 
constituting  a  tenfold  linear  en- 
8»  =  n  2  largement  of  AB. 

If  the  sines  of  the  angles  were 

given  by  intervals  of  ten  degrees,  interpolation  by  tenths  would 
give  the  sines  by  degrees  ;  the  circle  with  radius  100  mm.  (or 
100  twentieths  of  an  inch)  permits  the  sine  and  cosine  to  be 
read  to  two  places  accurately,  and  this  rather  low  degree  of 
refinement  corresponds  to  a  table  of  sines  given  by  intervals 
of  ten  degrees  ;  interpolation  would  give  substantially  correct 
values  to  two  decimal  places,  e.g.  for  the  sines  of  21°,  22°, 


-   0  1  2  3  4 n-ln 

Graphical  representation  of  the 
sum  of  an  arithmetical  series 


ARITHMETICAL  SERIES  AND  INTERPOLATION     171 


Arc  of  20°  to  30°  in  circles  with  radii  50,  100,  and  1000  fortieths  of  an  inch 

The  marks  on  the  long  chord  indicate  the  points  given  by  interpolating 
between  sin  20°  and  sin  30°,  and  between  cos  20°  and  cos  30°.  Even  on 
the  arc  with  25  inch  radius  nine  interpolated  points  on  the  chord  and  cor- 
responding points  on  the  arc,  between  21°aud  22°,  coincide. 


172  UNIFIED  MATHEMATICS 

...  29°  by  interpolating  between  sin  20°  =  .34  and  sin  30°  =  .50. 
The  arc  of  10°  on  this  circle  differs  slightly  but  appreciably  to 
the  eye  from  the  chord  of  10°,  but  the  interpolated  points  on 
the  chord  are  not  easily  distinguished  from  the  ten  points 
on  the  curve. 

Angles  given  by  degrees  permit  interpolation  by  intervals  of 
6'  or  by  intervals  of  10',  with  substantially  correct  values  to 
the  third  place  if  the  values  are  given  only  to  three  places  ; 
values  given  to  four  places  give  by  interpolation  values 
substantially  correct  to  the  fourth  place.  On  our  circle  with 
radius  1000  the  sine  and  cosine  can  be  read  to  three  decimal 
places  ;  interpolation  between  the  values  of  sin  20°  and  sin  30° 
give  points  markedly  different  from  the  true  points  on  the 
curve.  These  points  are  indicated  by  checks  on  the  chord  of 
10°.  Interpolating  five  points  (for  10',  20',  30',  40',  50')  on  the 
chord  from  the  20°  point  to  the  21°  point  gives  points  not 
readily  to  be  distinguished  from  the  correct  points  on  the  arc. 
On  this  diagram  it  is  not  possible  to  distinguish  the  sub- 
divisions for  minutes  on  the  arcs  from  the  corresponding 
points  on  the  chords  of  central  angles  of  10'. 

With  the  proper  changes,  noting  particularly  that  as  0  in- 
creases cosine  0  decreases,  the  argument  given  holds  'for  inter- 
polated values  for  cos  0. 

Interpolation  of  the  tangent  values  is  similar,  except  in  the 
neighborhood  of  90°  where  the  tangent  changes  very  rapidly ; 
in  a  separate  table  are  given  by  minutes,  the  tangents  of  angles 
from  88°  to  90°. 

The  graph  of  the  function  y  =  logw  x,  or  10"  =  x,  is  a  con- 
tinuous curve  which  for  small  arcs  approximates  a  straight 
line.  Similarly  the  graphs  of  the  functions  y  =  sin  x,  and  of 
2/=sin  x,  log  cos  x,  log  tan  x  and  log  cot  x  approximate  straight 
lines  within  small  intervals,  and  so  are  subject  to  our  ordinary 
process  of  interpolation. 

6.  Arithmetical  means.  —  If  two  numbers  a  and  b  are  given, 
the  arithmetical  mean  between  the  two  is  the  number  x  which 


ARITHMETICAL  SERIES  AND  INTERPOLATION     173 

makes  a,  x,  b  three  consecutive  terms  of  an  arithmetical  series  ; 
to  insert  n  arithmetical  means  it  is  necessary  that  a,  the  n 
means,  and  b  form  n  +  2  consecutive  terms  of  an  arithmetical 
series.  Ordinary  interpolation  is  the  insertion  between  two 
tabular  values  of  some  particular  one  of  9  arithmetical  means. 
If  a,  x,  b  form  an  arithmetical  series, 

b  —  x  =  x  —  a, 

a  +  b 
whence  x  =  —  ~  -- 

If  a,  a  -f-  d,  a  +  2  d,  a  -\-  3  d,  —  a  -\-(n  —  l)d,  a  -|-  nd,  b  form 

an  arithmetical  series,  b  =  a  +(n  +  l)d;  whence  d  =     ~  a  . 

n  +  1 
The   sum   of  n  terms   of   the  series   a,  a  -\-  d,  a+2d}   ••• 

a  +  (n  —  l)d,  is 


the  average  value  of  these  n  numbers  is 

n(a  +  £„)  _  a  -\-  ln 
2-n  2     ' 

•termed  the  arithmetical  mean  of  the  n  numbers  ;  the  sum  of 
an  arithmetical  series  is  seen  to  be  the  "  average  value  "  mul- 
tiplied by  the  number  of  terms.  Similarly  of  any  collection 
whatever  of  n  quantities,  the  arithmetical  mean  is  regarded  as 
the  total  sum  divided  by  the  number  of  quantities.  In  statisti- 
cal work  the  latter  mean,  total  sum  divided  by  the  number  of 
given  quantities,  is  called  the  "  weighted  mean." 

PROBLEMS 

1.  Between  .3584  and   .3746  insert  5  arithmetical  means; 
if  .3584  =  sin  21°  and  .3746  =  sin  22°,  what  do  these  means 
represent?     Between   .3746   and   .3584  insert  5  arithmetical 
means  ;  interpret  as  cosines. 

2.  Given   sin   21°  =  .3584   and  sin  21°  10'  =  .3611,  find  9 
intermediate  values  ;  interpret. 


174  UNIFIED  MATHEMATICS 

3.  Given  sin   0°  =  0,  sin  30°  =  .5000,  what  value  would 
arithmetical   interpolation  give   for   sin   21°?     What   is   the 
error  ? 

4.  Given  sin  20°  =  .3420  and  sin  30°  =  .5000 ;   find  to  4 
places  sin  21°.     How  many  terms  in  the  arithmetical  series 
which  is  implied  ? 

5.  What  is  the  sum  of  the  first  ten  integers  ? 

6.  If  cards  are  marked  1  to  190,  what  is  the  total  sum  ? 
What  is  the  average  value  of  the  total  group  of  numbers  ? 

7.  How  many  years  of  life  have  been  lived  by  a  group  of 
30  individuals,  aged  21,  22,  23,  ...  50  years? 

8.  Falling  from  rest  a  body  falls  approximately  16  feet  in 
the  first  second,  and  48  in  the  second,  and  in  each  succeeding 
second  32  feet  more  than  in  the  one  which  precedes.     What 
distance  will  the  body  fall  in  10  seconds  ?     How  long  will  it 
take  such  a  body  to  fall  1000  feet  ? 

9.  If  it  takes  a  lead  ball  8  seconds  to  fall  to  the  earth  from 
a  balloon,  what  is  the  height  of  the  balloon  ? 

10.  How  long   will    it   take  a  ball  to  reach  .the  earth  if 
dropped  from  the  top  of  the  Washington  monument,  550  feet 
high  ? 

11.  Draw  figures  to  show  that  between  x  =  3  and  x  =  4, 
the  graph  of  xy  =  1  approximates  a  straight  line. 

12.  Write  the  equation  of  a  straight  line  representing  for 
integral  values  of  x,  from  0  to  10,  the  arithmetical  series  with 
a  =  10,  d  =  —  3.     Represent  the  series  also  by  the  series  of 
rectangles,  each  of  width  1.     In   summing   the   arithmetical 
series  we  reversed  the  series  and  added  ;  show  the  geometrical 
equivalent  on  the  figure,  page  170,  with  the  rectangles. 

13.  Given  that  the  first  term  of  an  arithmetical  progression 
is  8  and  the  last  term  100,  what  equation  must  n  and  d  satisfy  ? 
If  d  =  2,  what  is  n  ?     If  d  =  3,  what  is  n  ?     Interpret. 


ARITHMETICAL  SERIES  AND  INTERPOLATION     175 

14.  Show   that   in    an  arithmetical  series  of  n  terms   (a, 
a  +  d,  •••)  the  average  value  is  ^  the  sum  of  the  first  and  last 
term.     Note  that  the  average  value  of  the  terms  is  the  total 
sum  divided  by  the  number  of  terms.     This  average  value  is 
termed  the  arithmetical  mean  of  the  n  terms. 

15.  Find  the  average  value  of  the  following  10  heights,  and 
the  arithmetical  mean : 

40  6  feet  126  5  feet  9£  inches 

64  5  feet  11^  inches  138  5  feet  9  inches 

86  5  feet  11  inches  120  5  feet  8£  inches 

92  5  feet  10|  inches  112  5  feet  8  inches 

142  5  feet  10  inches  80  5  feet  7J  inches 

16.  If  there  are  1000  men  measured  and  they  are  grouped 
in  height  as  above,  find  the   average  height  of  these   men. 
Note  that  the  easiest  way  to  find  this  average  is  to  take  the 
variation  above  and  below  some  one  of  the  "  middle "  values, 
e.g.  with  reference  to  5  feet  9  inches  as  origin,  6  feet  is  re- 
garded as  +  3  inches  and  this  group  has  a  total  of  120  inches 
excess  above  5  feet  9  inches  per  individual ;  5  feet  1\  inches  is 
regarded  —  1^  inches  and  the  total  group  of  80  has  a  total 
deficiency  of  120  inches,  or  —  120  inches ;  the  two  neutralize 
each  other.     Whatever  total  remains  is  divided  by  1000  and 
added,  algebraically,  to  the  5  feet  9  inches. 

17.  Draw  the  graph  of  I  =  a  +(n  —  l)d;  assuming  a  and  d 
as  constants. 

18.  Historical  problem.  —  In    the    Egyptian   manual   men- 
tioned above,  occurs  the  following  problem :  If  100  loaves  of 
bread  are  divided  according  to  the  terms  of  an  arithmetical 
series  among  5  people  so  that  ^-  of  what  the  first  three  receive 
equals  what  the  last  two  receive,  find  the  number  received  by 
each  person.      Solve  the  problem.      The   Egyptian  reckoner 
assumes  that  the  last  person  receives  1  loaf,  and  without  any 
explanation,  that  the  second  receives  6^  loaves,  and  so  on  in 
arithmetical  progression ;  the  sum  he  finds  to  be  60,  and  to 
arrive  at  the  correct  values  all  numbers  are  increased  in  the 
ratio  of  100  to  60.     Compare  with  your  solution. 


CHAPTER   XI 

GEOMETRICAL    SERIES   AND   APPLICATIONS   TO 
ANNUITIES 

1 .  Geometrical  series .  —  A  series  of  terms  in  which  each  term  is 
obtained  from  the  preceding  by  multiplying  by  a  fixed  number 
is  called  a  geometrical  series  ;  by  definition  the  ratio  of  each 
term  to  the  preceding  is  a  constant.     Designating  this  ratio 
by  r,  and  the  first  term  by  a,  the  type  series  becomes : 

a,  ar,  ar2,  ar3,  ar""1. 

ln  =  arn~l. 

The  sum  of  such  a  series  to  n  terms  is  obtained  as  follows : 
Let  sn  =  a  +  ar  +  ar2  +  ar3  +  ...   +  arn~l. 

rsn  =  ar  +  ar2  +  ar3  +   ...   -+-  ar""1  +  ar". 
sn  —  rsn  =  a  —  arn. 
sn(l  —  r)  =  a  —  ar". 

_  a  —  ar*  _     a  ar"    _  ar"  —  a 

1—r       1—r     1—r       r— 1 

2.  Sum  to  infinity,  r<  1.  —  When  r  is  numerically  less  than 
1,  the   terms  of    a    geometrical   series  become   smaller  and 
smaller  without  limit.     The  sum  of  n  terms  differs  from  the 

fixed  quantity,   ,  by  the  quantity  -  '- — ,  which  decreases 

1—r  1—r 

C1  Cl  ClVn 

as  n  increases  ;  this  difference  between  -     —  and  - 

1—r          1—r      1—r 

can   be   made  smaller  than  any  assigned   quantity   however 

small  by  taking  n  sufficiently  large  ;  the  value  -     -  is  termed 

1  —  r 

176 


GEOMETRICAL  SERIES  AND  ANNUITIES       177 

the  "  sum  to  infinity "  of  the  series  or,  more  strictly,  the 
"  limit "  of  the  sum  of  the  series  as  the  number  of  terms  in- 
creases indefinitely. 

A  simple  and  familiar  illustration  of  a  geometrical  "  sum  to 
infinity  "  is  found  in  the  recurring  decimals  of  elementary  arith- 
metic. Thus  .33333  ...,  or  .3,  is  the  series  T3¥  -fyf^  +  y-^  +  ... 
with  r  =  y1^,  a  =  T3¥  ;  .it  represents  the  series  TyT  +  y^y^ 

+  TTlWW  +   -  With  *  =  T*1F»  a  =  loV.   »  =  *  =  |j. 

Another  illustration  of  an  infinite  geometrical  series  is  found 
in  the  total  distance  traversed  in  passing  from  one  point  to 
another  by  passing  first  through  -|  of  the  distance,  then  through 
half  of  the  remaining  distance,  and  successively  in  each  sub- 
sequent movement  through  half  of  the  distance  which  remains 
to  the  goal.  One  of  the  famous  paradoxes  of  Zeno  is  to  the 
effect  that  the  hare  cannot  overtake  the  tortoise  since  the  hare 
must  first  cover  one  half  the  distance  intervening  between  their 
original  positions,  then  one  half  of  the  remaining  distance, 
then  one  half  the  remaining  distance,  and  in  each  subsequent 
interval  of  time  one  half  of  the  distance  which  remains ;  conse- 
quently the  hare  cannot  overtake  the  tortoise  as  by  this 

I 1 I TT~ 

A  8  Ml  4  Mt       1       M3\  M^lB 

Hare  Tortoise 

process  there  is  always  distance  intervening,  and  in  the  mean- 
time, further,  the  tortoise  has  advanced.  If  each  of  these 
intervals  of  space  of  the  infinite  series  traversed  required  the 
same  length  of  time,  or  any  finite  portion  of  time,  the  argu- 
ment would  be  sound,  but  you  have  here  two  infinite  series 
each  with  a  finite  sum.  As  each  space  interval  becomes 
smaller,  the  time  required  to  traverse  that  space  becomes 
smaller.  The  total  distance  evidently  is  the  sum  of  8  +  4  +  2 
+  1  -f  ^  +  •  •  •  which  has  the  sum  16,  both  by  the  formula  and 
by  the  figure ;  to  cover  the  whole  distance  requires  1  hr.  and 
this  takes  the  hare  up  to  B.  The  fallacy  in  Zeno's  argument 
lies  in  restricting  the  discussion  to  limited  intervals  of  space 


178  UNIFIED  MATHEMATICS 

and  time  preceding  the  instant  and  place  at  which  the  tortoise 
is  overtaken. 

3.   Geometrical  means.  —  If  a,  x,  b  form  a  geometrical  series 

b      x  — 

-  =  -,  whence  x2  =  ab,  x  =  Vab. 

x     a 

If  a,  ar,  ar2,  ar3,  ar4,  •««  ar""1,  arn,  b  form  a  geometrical  series, 

evidently 

b  =  arn  '  r  =  arn+l, 

Vfty* 


whence  r  =  (  -  ]     ,  and  the  series  is 
\aj 

1  2  _n  n+1 

a,  a(  -  )     ,  a.  -  )     ,  •••,  a(-\     ,  af  -  ]      or  6. 
\aj          \aj  \aj          \aj 


PROBLEMS 

1.  Sum  to  twenty  terms  the  series  2,  4,  8,  16,  •••. 

In  the  following  series  give  the  sum  to  twenty  terms,  and 
where  possible  give  the  sum  "  to  infinity." 

2.  3,  -  6,  +  12,  -  24,  +  48,  .... 
^^11      i     ... 

d-     ^5   2>  T2>  72>        ' 

4.  .1717171717  — ,  or  .17  (repeating  decimal). 

5.  .01717171717  •-,  or  .017. 

6.  3.16161616  .... 

7.  5,  15,  45,  135,  »•. 

8.  4,  10,  28,  82,  244,  ...  the  wth  term  being  1  +  3". 

9.  3,  7,  11,  15,  19,  .... 

10.  3,  —  7,  —  17,  -  27,  •». 

11.  v,  u2,  -y3,  v*,  ....     Sum  to  n  terms. 

12      1     r    r2    r3    r4    .. 
AX<.    ±,  /,/,/,/, 

13.  (1  +  i),  (1  +  i)*>  C1  +  O3?  (1  +  0*1  '"•     Sum  to  n  terms. 

14.  The  number  of  direct  ancestors  which  an  individual  has 
is  represented  by  the  series,  2, 4,  8,  16,  — .     Find  the  total  num- 
ber in  the  preceding  10  and  in  the  preceding  20  generations. 


GEOMETRICAL  SERIES  AND  ANNUITIES       179 

15.  According  to  Galton's  law  of  heredity  the  parents  con- 
tribute to  the  hereditary  make-up  of  an  individual  -|-  of  what  is 
contributed  by  all  the  ancestors ;  the  grandparents  contribute 
^ ;  the  great-grandparents,  of  whom  there  are  eight,  contribute 
^  ;  and  so  on  ;  find  the  total  contribution.    Find  the  individual 
contribution  of  a  single  individual  four  generations  back. 

16.  Insert  three  geometric  means  between  2  and  17 ;  com- 
pute to  one  decimal  place. 

17.  Insert  one,  two,  and  three  geometric  means  between 
1  and  2. 

18.  One  of  the  so-called  "  three  famous   problems  of  an- 
tiquity "  is  to  construct,  using  only  ruler  and  compass,  a  cube 
which  is  double  the  volume  of  a  given  cube.    This  problem  was 
very  soon  reduced  to  the  insertion  of  two  geometric  means  be- 
tween a3  and  2  a3 ;   insert  two  geometric  means  between  a3 
and  2  a3  and  show  that  this  gives  the  algebraic  value  of  the 
side  of  a  cube  of  double  the  volume ;  the  geometrical  solution 
has  been  demonstrated  to  be  impossible  if  ruler  and  compass 
are  the  only  instruments  of  construction. 

19.  In  the  population  statistics  on  page  65  find  the  population 
of  the  United  States  in  1810  and  1910 ;  between  these  two 
numbers  insert  9  geometric  means  ;  find  r ;  this  represents  the 
approximate  decennial  rate  of  increase  in  the  population  at 
1810  which  would  give  the  final  actual  population  in  1910. 
Compare  with  the  actual  census  figures  at  the  end  of  each  ten 
years. 

20.  How  would  you  find  the  regular  annual  rate  of  increase 
in  the  population  of  the  United  States,  in  other  words  the  fixed 
annual  percentage  increase  in  population  which  would  change 
the  population  from  7.2  millions  in  1810  to  101.1  millions  in 
1910  ? 

21.  The  plunger  chamber  of  an  air  pump  is  approximately 
JT  of  the  total  air  capacity ;  at  each  stroke  ^  of  the  air  in  the 
receiver  is  removed ;    after  10,  15,  and  20   strokes  find  the 


180 


UNIFIED  MATHEMATICS 


proportionate  amount  of  air  ivmiiiiiiiig  in  the  receiver ;  approx- 
imately how  many  strokes  must  be  made  to  remove  99%  of 
the  original  air  ? 

4.  Graphical  representation  —  Geometrical  series.  —  On  the 
straight  line  y  —  rx  of  slope  angle  «,  with  tan  a  =  r,  if  x  =  a, 
ar,  ar2,  ar3,  —  the  successive  abscissas  represent  the  terms  of 


Graphical  summation  of  the  ge&metrical  series,  r  <  1 
Note  that  the  "  sum  to  infinity,"  r  <  1,  is  represented. 


the  geometrical  series ;  by  means  of  the  45°  line  through  the 
origin  these  successive  abscissas  are  readily  constructed. 
However,  a  more  favorable  construction  for  a  graphical  treat- 


GEOMETRICAL  SERIES  AND  ANNUITIES       181 

ment  of  the  series  regards  each  term  of  the  series  a,  ar,  ar2, 
ar3,  —  as  an  addition  to  the  preceding  abscissa;  the  succes- 
sive additions  to  the  ordinates,  increments  of  the  ordinates, 
will  be  ar2,  ar3,  a?*4,  — .  By  drawing  the  line, 

y  =  x  —  a 

at  an  angle  of  45°  with  the  #-axis,  the  successive  abscissas  are 
readily  constructed ;  by  drawing  through  the  point  Pi(a,  ar)  a 
line  parallel  to  the  or-axis  it  intersects  the  45°  line  drawn 
through  (a,  0)  at  a  point  M2  such  that  1\MZ  =  PiAly  since 
Z.  PiAiM2  =  45° ;  a  parallel  to  the  y-axis  through  M2  intersects 
the  line  y  =  rx  at  a  point  P2  such  that  M2P2  =  r  •  P\M*  =  ar2, 
and  A2P2  =  ar  +  a?'2.  Similarly,  if  Pn  represents  the  wth 
point  found  on  our  line,  y  =  rx, 

MnPn  =  arn,  OAn  =  sn  =  a  +  ar  -f-  ar2  -f-  •••  ar""1, 

P  A 

— = — £  =  tan  a  =  r,  whence  PnAn=  rsn ;  AnMn=  srn  —  ar" ;  A±An 

=  sn  —  a ;    but    AlAn  =  AnMH,  .since    the   slope    of    the   line 
is  1. 

.•.  sn  —  a  =  rsn  —  ar", 

sn(l  —  ?•)  =  a  —  ar", 
a  —  ar" 


«„  = 


1-r 


If  r  <  1,  as  in  our  figure  1,  a  <  45°,  and  the  two  lines  inter- 
sect at  K  to  the  right  of  the  origin  ;  and  the  points  of  intersec- 
tion M2,  Ms,  —  Mn  •••  will  fall  below  K  on  the  line  BK.  The 
abscissa  of  K  represents  the  "  sum  to  infinity  "  of  our  series,  as 
it  is  evident  that  the  series  of  triangles  could  be  continued 
indefinitely  in  the  opening  OKA.  Evidently  also,  solving 
y  =  x  —  a,  y  =  rx, 

rx  =  x  —  a 


^ 

1  —  r 


182 


UNIFIED   MATHEMATICS 


Graphical  summation  of  the  geometrical  series,  r  >  1 

For  r  >  1,  the  figure  is  quite  similar ;  the  two  lines,  y  =  rx 
and  y  =  x  —  a,  diverge  ;  tan  «  =  r,  a  >  45°. 
Evidently,  AnMn  =  A:An, 

An-Ln  :=  ^*^BJ 

AnMn  =  rsn  —  arn, 
OAn  =  s. 
AtAn  =  s  —  a, 

whence  sn  —  a  =  ?-sn  —  ar". 

sB(l  —  r)  =  a  —  ar", 

as  before  =.""  ar"  or  art>~a. 

1-r  r-1 


GEOMETRICAL  SERIES  AND  ANNUITIES       183 

5.  Historical    note.  —  Arithmetical    and    geometrical    series 
are  found  in  the  oldest  mathematical  documents  known,  both 
in  the  remains  of  ancient  Egypt  and  of  ancient  Babylon.     The 
system  of  numbers  used  by  the  Babylonians  as  early  as  2000- 
3000   B.C.    was   sexagesimal,  increasing  by  powers   of   60   in 
geometrical  series.     Further  an  early  Babylonian  clay  tablet 
gives  the  portion  of  the  moon's  surface  illuminated  on  each  of 
fifteen  successive  nights,  from  new  moon   to  full  moon  by  a 
geometric  and  an  arithmetical  series.     The  moon's  surface  is 
conceived  as  divided  into  240  parts  ;  on  the  first  five  nights 
5,  10,  20,  40,  and  80  parts,  respectively,  are  illuminated  and 
011   the   following   ten  nights,  96,  112,    128,   144,  •••  and   on 
in  arithmetical  progression  to  240. 

The  Egyptian  manual  of  mathematics  of  1700  B.C.  (or  there- 
abouts), includes  two  rather  complicated  problems  on  arith- 
metical series,  involving  also  the  insertion  of  means,  and  one 
problem  involving  the  summation  of  a  geometric  series. 

The  equivalent  of  a  general  formula  for  summation  of  a 
geometrical  series  was  first  established  rigorously  by  the 
Greeks,  and  appears  in  Euclid's  Elements,  Book  IX,  prop.  36. 
The  first  summation  "  to  infinity  "  of  a  decreasing  geometrical 
progression  was  effected  by  the  great  Archimedes  (287-212 
B.C.)  who  employed  the  formula  in  finding  the  area  of  a  seg- 
ment of  a  parabola. 

In  a  great  part  of  the  later  development  of  mathematics 
such  series  have  played  a  prominent  role,  in  some  measure 
because  of  their  own  intrinsic  importance  and  in  some  measure 
as  fundamental  in  the  discussion  of  other  types  of  series. 

6.  Annuity  formulas,     a.   Accumulated  value  of  an  annuity. — 
The  geometrical  series  plays  a  large  role   in   the   theory   of 
investments  and  insurance.     We  have  shown  (p.  54)  that  at 
rate  i  per  year,  compounded  annually,  1  will  amount  in  n  years 
to  (1-f  /)" ;  if  1  is  invested  at  rate  i  at  the  end  of  each  year,  i.e. 
annually,  for  n  years,  these  payments  constitute  an  annuity  of 
1  for  ?2  intervals,  at  i  per  annum.     The   total   accumulated 


184  UNIFIED   MATHEMATICS 

value  SJPJ,  at  the  end  of  n  years  of  such  an  annuity,  is  the  sum 
of  the  geometrical  series 

(i  +  t)-i  (i  +  o-2>  (i  +  0-'»  (i  +  O-4,  •-  (i  +  0» !» 

since  the  first  payment  made  at  the  end  of  the  first  year  ac- 
cumulates for  (n  —  1)  years,  the  second  for  (n— 2)  years,  ••-, 
and  the  w.th  payment  of  1  is  made  at  the  end  of  the  n  years. 
The  sum,  called  the  amount  of  the  annuity, 


represents  the  accumulated  value  of  an  annuity  of  1  per  inter- 
val for  n  years  or  intervals  at  a  rate  of  i  per  year  or  interval. 

b.    The  annuity  ichich  will  accumulate  to  1.  —  Very  evidently 
K    per     annum    will     produce     at     the    end    of     n    years 

/|    _j_  j'Nn   ^ 

Ksn  =  K  •  * —  —  ;    the  annuity   which   in  n  years   will 

i 

amount  to  1  is  evidently  the  value  of  K  which  makes  Ksn  =  1 ; 
this  value  is  —  = • 


c.  Present  value  of  an  annuity  of  1. — The  accumulated 
value  of  1  to  be  paid  at  the  end  of  n  years  at  i  per  year 
is  (1  +  i)n ;  K  will  accumulate  at  the  rate  t  in  n  years  to 
K(l  +  ?')";  this  means  that  K  dollars  (or  units)  in  hand  ac- 
cumulates to  JT(1  +  i)n  dollars,  and  that  K(l  +  £)"  dollars 
to  be  paid  n  years  hence  is  worth  K  dollars  now,  money  at 
rate  i  per  year.  Hence  the  present  value  of  1  to  be  paid  n 
years  hence  is  a  value  of  K  which  makes  /i"(l  +  i)n  =  1,  or 

K  = =  y",  wherein  v  = The   present   value   of 

(i  +  0"  i  +  * 

an  annuity  of  1  per  annum  for  n  years  when  money  is  worth 
i  per  annum  is  the  sum  of  the  geometrical  series, 

v,  v2,  v3,  v4,  v5,  i£,  i:7,  —  vn. 

The  first  term  v  is  the  present  value  of  the  first  payment  of  1 
which  is  to  be  made  1  year  from  date ;  the  second  term  is  the 


GEOMETRICAL  SERIES  AND  ANNUITIES       185 

present  value  of  the  second  payment  of  1  to  be  made  in  2 
years ;  ••• ;  vn  is  the  present  value  of  the  final  payment  of  1  to 
be  made  n  years  hence. 


v  —  vn+l      1  —  vn      1  — V 


d.  TJie  annuity  which  1  will  purchase.  —  The  present  value  of 
K  per  annum  for  K  years  is  Ka^\  =  K — —. — ;  the  annuity 

v 

which  is  worth  1  at  the  present  time  is  evidently  a  value  K 

which  makes  KatTl  =  1,  whence  K=  —  = •     This  is  the 

annuity  which  1  will  purchase. 

e.  Summary  of  interest  functions.  — 
These  six  functions, 

rn  —  (1  -f  £)*,  accumulation  of  1, 
vn  =  (1  4-  j)~",  discount  value, 

s^n  =  ^ —      '   ~ — ,  accumulated  annuity  value, 
i 

1  —  (1  4-  rrn      1  —  vn 

a^i  = • * —  =  —     — ,  present  value  of  annuity, 

i  i 

,  annuity  to  accumulate  to  1, 


«ti    (1  +  0" 

1  i 

and  —  =  —  — ,  the  annuity  which  1  will  purchase, 

a,r|     1  —  (1  4-  0  " 

are  of  fundamental  importance  in  the  valuation  of  bonds,  and 
in  all  problems  where  stipulated  payments  are  to  be  made  at 
stipulated  intervals,  and  also  in  the  theory  of  interest. 

If  interest  is  to  be  compounded  semiannually  and  pay- 
ments are  made  semiannually,  the  interval  can  be  considered 
as  6  months  and  the  rate  of  interest  as  one  half  the  stated 
rate ;  similarly  the  interval  can  be  considered  as  3  months  and 
the  rate  of  interest  as  one  fourth  the  stated  interest  if  interest 


186  UNIFIED  MATHEMATICS 

is  compounded  quarterly  and  payments  made  quarterly. 
Other  types  of  problems  with  payments  falling  between  in- 
terest periods  are  beyond  the  scope  of  this  work. 

PROBLEMS 

1.  Compute  the  value  of  s     f or  &%,  5%,  4%,  using  loga- 
rithms to  obtain  (1  +  .06)20,  (i  +  .05)20,  and  (1  +  ,04)w.     What 
percentage  of  error  is  introduced  using  four-place  tables  ?     Dis- 
cuss the  effect  in  finding  the  accumulated  value  of  an  annuity 
of  $  100  per  year  for  20  years.     Check  by  the  tables  given  at 
the  back  of  this  book. 

2.  Compute  a—,,  and  discuss  as  in  1. 

3.  Find  by  logarithms  from  your  values  of  s^  and  a^-,  the 
annuity   which   will   accumulate   to   1    in  20  years,  and  the 
annuity  which  1  in  hand  will  purchase. 

4.  If  payments  of   $50  per  year  are  made  semiannually 
and  interest  is  compounded   semiannually,  find  the  accumu- 
lated  value  of   this   annuity  at   the   end  of   20  years  for  a 
nominal  interest  rate  of  6  % ,  5  % ,  and  4  %  respectively,  per 
annum  (3  %,  2^  %,  and  2  %  per  interval).     Use  the  tables. 

5.  What  annual  payments  continued  for  10  years  are  equal 
to  $1000  cash  in  hand? 

6.  What  annual  payments  continued  for  10  years  are  equiv- 
alent to  $  1000  to  be  paid  at  the  end  of  10  years?  to  $  1000  to 
be  paid  at  the  end  of  20  years? 

7.  Prove  8^=  (1  + i)"^;  and  a^sw"- Sg-].     Discuss. 

8.  Show  algebraically  that =  i. 

a,7]     s«\ 

NOTE.  —  The  difference  between  1  in  hand  and  1  to  be  paid  in  n  years 
is  simply  the  earning  power  of  the  1  in  hand  for  this  period  of  n  years ; 
if  money  is  worth  6  %  per  year,  1  in  hand  will  earn  every  year  for  n 
years  .06  in  addition  to  preserving  itself  ;  1  in  n  years  is  worth  simply  1 
then ;  —  is  the  annuity  for  n  years  which  1  in  hand  purchases,  and  —  is 


GEOMETRICAL  SERIES  AND  ANNUITIES       187 

the  annuity  equivalent  to  1  to  be  paid  in  n  years  ;  the  difference  is  the 
annual  earning  of  the  1  in  hand,  i. 

9.  Give  the  arithmetical  series  represented  by  the  ordinates 
of  y  =  3  x  +  7,  for  integral  values  of  x  from  0  to  10.  What  is 
the  sum  of  this  series  ?  Between  1  and  2  interpolate  9  values, 
at  equal  intervals,  and  state  corresponding  series  ;  what  corre- 
sponds to  the  tabular  difference  ? 

10.   Discuss  as  in  problem  9  the  corresponding  ordinates  of 


11.  Sum  to  20  terms  the  series  7,  5,  3,  1,  —  1,  —  . 

12.  Write  the  20th  term  of  7,  5,  3,  1,  —  1,  .... 

13.  Write  the  tenth  term  of  7,  5,  ^5-,  -L2/,  .... 

14.  Find  the  arithmetical  and  the  geometrical  means  be- 
tween 7  and  5. 

15.  Historical   problem.  —  It    is    related    that    an    Indian 
prince  who  wished   to  reward   the   inventor  of   the  game  of 
chess    suggested  to   the  inventor  that   he   should  name    the 
reward  he  desired.     The  scholar  replied  that  he  would  take 
1   grain   of  wheat  for  the  first   square   of   the   board,  2    for 
the  second,  4  for  the  third,  8  for  the  fourth,  16  for  the  fifth, 
and  so  on  in  geometrical  progression  to  cover  the  64  squares. 
The  prince  agreed  but   found,  on  the  computation,  that  the 
value  exceeded  that  of  his  realm.     Taking  10,000   grains  as 
approximately  a  pint,  make  a  rough  calculation  of  the  amount 
involved. 

16.  Historical   problem.  —  In    textbooks    of    the    sixteenth 
century  the  following  problem  frequently  appears.     A  black- 
smith being  asked  his  price  for  shoeing  a  horse  replied  that 
for  the  first  nail  he  would  charge  one  fourth  of  one  cent  (use 
this  in  place  of  farthing,  or  pfennig),  £  cent  for  the  second 
nail,  1  cent  for  the  third,  2  for  the  fourth,  and  so  on  for  the 
thirty-two  nails.     Compute  the  price. 


188  UNIFIED  MATHEMATICS 

7.  Annuity  applications.  —  Brief  tables  of  the  annuity  func- 
tions are  given  at  the  back  of  this  book ;  somewhat  larger 
tables  will  be  found  in  the  Bulletin  No.  136  of  the  U.  S. 
Bureau  of  Agriculture,  which  includes  also  a  more  extensive 
treatment  of  the  subject  of  bonds  and  annuities  by  Professor 
James  W.  Glover. 

a.  Common  annuity.  —  To   find   the  purchase  price  of  an 
annuity  of  k  dollars  per  interval  for  71  intervals  when  the  current 
rate  is  i  per  interval  is  obviously  a  direct  application  of  the 
a;n  table,  as  the  purchase  price  is  simply  the  present  value  of 
the  series  of  payments.     If  the  first  payment  of  the  annuity  is 
to  be  made  r  intervals  hence,  a  deferred  annuity  for  n  intervals, 
the  price   may  be  considered   as  the  difference  between  an 
annuity  for  r  —  1  intervals  and  an  annuity  for  n  +  r  —  1  inter- 
vals,    fln+r-i!  gives  a  payment   every   interval   for   n  +  r  —  1 
intervals,  the  first  made  at  one  interval  from  the  present  time ; 
af=ri  gives  a  payment  every  interval  for  (r  —  1)  intervals,  the 
first  as  before,  and  the  last  (r  —  1)  intervals  from  the  present 
time ;  the  difference  is  the  value  of  the  deferred  annuity  of  n 
payments,  first  payment  to  be  made  at  the  end  of  r  intervals. 

Some  large  banks,  trust  companies,  and  insurance  companies 
do  this  type  of  business.  Frequently  a  purchaser  desires  an 
annuity  to  be  paid  annually  terminating  with  the  death  of  the 
purchaser ;  this  involves  then  a  life  contingency,  and  the  dis- 
cussion and  solution  of  this  problem  require  new  methods  and 
new  tables. 

b.  Farm  loans.  —  To  extinguish  or  amortize  a   debt  by  n 
annual  payments  of  fixed  amount  is  the  type  of  problem  which 
arises   under   the   recent    Farm   Loan   Act.     Thus   a   farmer 
borrowing  $  10,000  at  the  bank  at  5  %  interest  may  desire  to 
make  such  a  payment  as  to  extinguish  the  debt  in  30  years, 
money  being   worth   5  %  '  annually.      The    problem   may  be 
solved  by  considering  the  annuity  which  $  10,000  will  purchase 
at    5  %    interest   for  30   years   or   $  10,000  x  .06505,   giving 
$  650.50.     The  problem  may  also  be  solved  by  considering  the 
interest  as  paid  each   year,   $  500,  in   addition   to  which  an 


GEOMETRICAL  SERIES  AND  ANNUITIES       189 

annual  payment  must  be  made  to  accumulate  at  5  %  to  $  10,000 
at  the  end  of  30  years.  This  annual  payment  is  found  to  be 
$  150.50.  The  value  found,  $  650.50,  may  be  checked,  as  below, 
by  using  the  $$  table.  Or  another  check  is  to  find  the  value 
at  the  end  of  30  years  of  the  $  10,000  or  $  10,000  (1  +  .05)30  and 
compare  this  with  the  value  of  $  650.50  x  %,. 

Suppose,  on  the  other  hand,  that  the  borrower  desires  to 
pay  approximately  $  600  per  year,  applying  the  extra  amount 

each  year  to  the  debt.    The  annuity  which  1  will  purchase,  —  5 

%f! 
is  the  function  involved.     The  question  here  may  then  be  put, 

For  what  period  of  years  at  5  %  interest  will  $  10,000  purchase 
an  annuity  of  $  600  per  annum  ?  We  will  consider  only  ap- 
proximate solutions,  taken  from  the  tables. 

The  tables  show  that  at  5  %  interest  $  10,000  will  purchase 

an  annuity  of  $  10,000  X  —  =  $  10,000  X  .06043  for  36  years, 

«36l 

or  $  604.30  annually  for  36  years ;  $  10,000  will  purchase  an 
annuity  of  $  598.40  for  37  years.  36  years  would  be  taken, 
and  this  period  of  36  years  is  provided  for,  as  an  amortization 
term,  by  the  government.  By  paying  every  year  $  100  more 
than  the  interest,  at  the  end  of  36  years  the  accumulated 
value  of  this  annuity,  the  excess  $  100  over  the  interest,  would 
be  worth  at  5  %  :  $  100  x  SMI  =  $  100  x  95.8363  =  $  9,583.63, 
lea,ving  $416.37  due  at  the  end  of  36  years.  This  amount 
with  interest  at  5  %  should  be  the  next  and  final  payment. 

c.  Sinking  funds.  —  If  a  city  issues  bonds  to  be  redeemed  20 
or  30  or  40  or  n  years  hence,  it  is  commonly  desirable  to  pro- 
vide for  the  repayment  of  the  bonds  by  an  annual  (interval) 
payment  allowed  to  accumulate  at  i  per  annum  (per  interval). 
Similarly  in  business  a  manufacturing  concern  using  an  ex- 
pensive piece  of  machinery  which  has  a  probable  lifetime  of 
20,  30,  or  40  years  must  provide  for  the  eventual  replacement 
of  this  machine  by  an  annual  payment,  out  of  earnings,  into  a 
sinking  fund.  In  this  type  of  problem  the  function  involved 
is  the  annuity  which  will  accumulate  to  1  in  n  years.  Thus 


190  UNIFIED  MATHEMATICS 

to  provide  for  replacement  of  a  $  10,000  piece  of  machinery 
in  30  years  money  at  5  %  requires   an   annual    payment  of 

$  10,000  x  —  or  S  10,000  x  -  -  which  equals  $  150.50 

5g6]  (1  +   I')30  -  1 

per  annum. 

d.  Bonds  at  premium  and  discount.  —  If  a  city  issues  bonds 
at  5  %  when  money  is  worth  in  the  money  market  4  %,  the 
bonds  will  sell  at  higher  than  face  value,  since  they  pay  on 
each  $  100  an  annuity  of  S  5.00  per  year  for  the  term  of  the 
bond,  when  investors  are  demanding  only  S  4.00  per  annum 
with  security  of  capital.  This  higher  price  is,  on  the  basis  of 
money  at  4  %,  the  present  value  of  an  annuity  of  $  1.00  per 
annum  for  the  term  of  the  bond  or  1  x  a^  at  4  % .  The  dif- 
ference between  the  par  value  or  face  value  of  a  bond  and  the 
price  offered  by  investors  is  called  the  premium  (or  discount, 
when  the  price  offered  is  less)  on  the  bond. 

If  a  city  issues  bonds  at  5  %  when  investors  are  demanding 
6  %,  a  bond  for  $  100  will  sell  at  a  discount  of  1  X  a^  at  6  %, 
since  the  investor  receives  from  these  bonds  not  S  6.00  per 
annum  but  only  $  5.00  per  annum.  Evidently  the  longer  the 
bond  has  to  run  the  greater  would  be  the  discount. 

In  general  terms  the  premium  on  a  bond  of  face  value  C, 
paying  a  dividend  rate  g,  bought  to  yield  j  per  annum  is 

P  =  C(g  —  j)a^  at  j  per  annum. 

PROBLEMS 

1.  If  a  father  sets  aside  annually  $  100  per  year  as  a  fund 
for  his  son  when  the  latter  becomes  of  age,  to  what  will  the 
fund  amount  at  the  end  of  21  years,  the  money  accumulating 
at  4  %  interest  ? 

2.  If  a  farm  mortgage  of  SI 0,000  draws  5  %  interest  and 
the  farmer  pays  annually  S  600,  what  is  the  accumulated  value 
at  the  end  of  21  years  of  the  excess  payments  of  S 100  per 
annum,  accumulated  at  5  %  ? 


GEOMETRICAL  SERIES  AND  ANNUITIES       191 

3.  What  annual  payment  will  accumulate  at  5  %  in  30  years 
to  $  10,000  ?     What  annual  payment  would  have  to  be  made 
on  a  $10,000  mortgage,  to  extinguish  the  debt  in  30  years, 
money  worth  6  %  ? 

4.  Find  in   the   tables   the   annuity   for  30  years  which 
$  10,000  will  purchase. 

5.  What    semiannual    payment    will    accumulate    in    30 
years    (60   payments)    to    $  10,000,   interest   being   5  %  com- 
pounded semiannually  ? 

6.  Find  the  cost  of  an  annuity  of  $  10,000  per  year  to  run 
for  10  years,  20  years,  and  30  years,  respectively,  money  being 
worth  4  %. 

7.  If  a  city  issues  $  10,000  in  bonds,  what  amount  must 
be  set  aside  annually  to  accumulate  at  4  °J0  interest  to  redeem 
t^e  bonds  at  the  end  of  20  years  ? 

8.  Find  the  cost  of  an  annuity  of  $  500  per  annum  for  10 
years,  the  first  payment  to  be  made  10  years  hence,  20  years 
hence,  and  30  years  hence,  respectively. 

9.  WThat   premium  can  you  afford  to  pay  on   a   $  10,000 
bond  drawing  5  %  to  run  20  years,  if  money  is  worth  4  °/0  ? 
What  discount  should  you  receive  if  money  is  worth  6  °/0  ? 

10.  What  is  the   present  value  of    $10,000  to  be  paid  20 
years  hence,  money  at  5  %  ? 

11.  Wrhich  is  the  better  offer  for  a  piece  of  property,  money 
being  worth  5  %,  a  rental  of  $600  per  year  for  20  years,  or  a 
price  of  $10,000?     A  rental  of  $600  per  year  for  10  years, 
and  $700  per  year  for  the  following  10  years,  or  $12,500? 
Assume  no  change  in  the  price  of  the  -real  estate  in  20  years. 

12.  What  sum  at  4  %  interest  should  a  railroad  set  aside 
each  year  to  replace  engines  worth  $35,000,  which  have  an 
estimated    life    of    25    years?    to    replace    buildings   worth 
$  1,000,000  which  have  an  estimated  life  of  100  years  ? 


192  UNIFIED  MATHEMATICS 

13.  If  a  man  invests    6100  each   year  for  20  years,  what 
annuity,  for  20  years,  can  he  purchase  at  the  end  of  the  first 
20  years,  money  at  5  %  interest  ? 

14.  If  a  man  agrees  to  take  $  1000  a  year  for  five  years  for 
a  house  originally  offered  at  S  5000,  what  is  the  discount  when 
money  is  worth  5  %  ? 

15.  At  5  %  interest  what  annual  payment  for  five  years  is 
equivalent  to  $  5000  cash  in  hand  ? 

16.  Answer  questions  14  and  15,  assuming   that  the  first 
$  1000  is  to  be  paid  immediately. 


1.  Binomial  series.  —  The  expressions  (1  +  i)4,  (1  -f  i)20,  — 
can  be  developed  in  powers  of  i  by  means  of  the  binomial 
expansion, 

( 


In  particular,  if  a  =  1, 


--  - 
1  1  •  ^  i  •  Z  •  o 

n(n  —  l)(n  —  2)(n  —  3)  to  (r  —  1)  factors     r_1 
1.2.3.4  ...  (r  -  1) 

This  formula  expresses  a  rule  for  the  formation  of  successive 
terms  of  the  expansion  of  (a  +  x)n  or  (1  +  x)n  ;  the  first  term 
contains  a  with  the  exponent  n  of  the  binomial  ;  the  second 
term  has  as  coefficient  the  exponent,  or  index,  of  the  binomial, 
x  appears  to  the  first  power,  and  the  exponent  of  a  decreases 
by  1  ;  the  coefficient  of  the  third  term  has  two  factors  in 
numerator  and  denominator,  in  the  numerator  n(n  —  1)  and 
in  the  denominator  1  •  2  ;  a  appears  with  exponent  1  less 
than  in  the  preceding  term,  and  x  with  exponent  1  greater  ; 
each  following  term  can  be  obtained  from  the  preceding  by 
introducing  one  further  factor  in  numerator  and  denominator 
and  at  the  same  time  decreasing  the  power  of  a  by  one  and 
increasing  that  of  x  by  one  ;  the  further  factor  in  the  new 
numerator  is  one  less  than  the  last  one  introduced  there,  and 

193 


194  UNIFIED  MATHEMATICS 

the  further  factor  in  the  new  denominator  is  one  greater  than 
the  last  one  of  the  preceding  denominator ;  the  coefficient  of 
the  term  in  x"~l  contains  in  the  numerator  (r—  1)  integral  factors 
from  n  down,  and  in  the  denominator  (r  —  1)  integral  factors 
from  1  up ;  x  appears  with  exponent  r  —  1  and  a  with  the 
exponent  which  added  to  r  —  1  makes  n,  i.e.  n  —  r  + 1. 

Illustrations  of  the  binomial  expansion. 

(\o  Oit5         n  ,      O    •    £  «       ,      O    •    ^    •    J.         Q 

a  +  x)3  =  a3  +  -  a*x  +  — —  ax*  +  x3 

=  a3  +  3  a2x  +  3  ax2  +  x3. 


IST  2o  8n 

6.  (a  +  a)15  =  a15  +  —  aux  +  ^— ^ 

lOrif  TERM 

15  •  14  . 13  . 12  . 11 . 10  .  9  •  8  •  7 
1.2.3.4.5.6.7.8.9 

Note  that  it  is  well  in  writing  the  tenth  term  to  begin  with 
xg ;  then  a  enters  to  the  sixth  power  as  in  every  term  of  this 
expansion  the  exponents  of  a  and  x  together  make  15 ;  the 
denominator  contains  9  factors,  1,  2,  3,  «••  9;  the  numerator 
contains  9  factors,  which  should  be  counted  as  they  are  writ- 
ten ;  finally  cancellation  should  be  made,  giving 

5  •  7  •  11  •  13  aW. 

15  - 14  . 13 


IOTH  TERM 

_  5  •  7  •  11  •  13  asx9  +  .... 


10TH 


Note  that  the  powers  of  a  in  each  term  can  be  dropped,  as 
every  power  of  one  equals  one. 


BINOMIAL  SERIES  AND  APPLICATIONS         195 

e.   Compute  to  4  decimal  places  (1  +  .04)15. 

IK.      1A  1 K      1A       1Q 

.  +  .04)16  =  1  +  15(.04) 


1-2   v  1-2-3 

15-14.13.12  15.14.13-12-11. Q 

1.2.3.4     l'  1.2.3.4.5 

.60  second  .0291  1.00000  first  term 

_1_  .12  .60000  second  term 

4.20  5).OQ349  fifth  .16800  third  term 

.04  .00070                                 .02912  fourth  term 


3).168Q  third  term         11  .00349  fifth  term 

.0560  .0077  .00031  sixth  term 

.04  .04  1.80092  Ans. 


.002240  .00031  sixth 

.02240  10  times 
672    3  times 
.02912  fourth  term,  to  be  multiplied  by  .12 

Note  here  the  method  of  computation  given  at  the  left ;  each 
term  is  obtained  from  the  preceding  term ;  three  new  factors 
of  which  one  is  .04  enter  into  each  succeeding  term,  two  in  the 
numerator  and  one  in  the  denominator ;  these  three  factors  after 

the  second  term  (.60)  are  14  X  '°4  or  7  x  .04 ;    then  13  X  -04; 

2  o 

then  12  x  -04  or   .12:   then  U  x  -04    which   might  well   be 
4  5 

treated  as  .008  x  11 ;  then  the  factor  — — —  would  give  the 

6 

seventh  term  from  the  sixth,  making  about  2  in  the  fifth  deci- 
mal place. 

The  expansion  of 


may  be  written  as  follows  : 


196  UNIFIED  MATHEMATICS 

in  which  T2,  T3,  Tt,  Tb,  •••  designate  the  second,  third,  fourth,  ••• 
terms  respectively.  This  type  of  representation,  in  which 
each  term  is  obtained  from  the  preceding,  is  frequently  of  use 
in  statistical  work  and  in  computation. 

/.  Write  the  sixth  and  sixteenth  terms  of  (1  -f-  #)". 

6-rii  TERM  16™  TEEM 

17-16.15.14.13  17-16... 

1.2.3.4.5  1.2... 

g.  What  decimal  place   is  affected   by   the   sixth   term  of 
(1.06)17  ? 

17 . 16  •  15  . 14  •  13 


1.2.3.4.5 


(.06)5;    (.06)3=  .000216  ;    (.06)4  =  .00001296 ; 


(.06)5  =  .00000078-  ;  multiply  .00000078  by  4,  this  by  7,  this 
by  13,  and  then  by  17,  rejecting  any  beyond  3  significant  fig- 
ures ;  this  gives  .00000312,  .0000218,  .000286,  and  finally  .00476. 
Note  that  the  computation  of  six  terms  of  (1  +  .06)17  involves  not 
very  much  more  numerical  labor  than  this  determination. 

h.  Write  six  terms  of  (a  — 


-  12  a"  (3  x)  +  ^~  «10(3  *)»  -  ™^'.S*  ^  ^ 

12.11,1019  _  12.11.10-9.8 

1.2.3.4  1.2.3.4.5 


ai2  _  36  a»a?  +  2  -  33  .  11  a10*2  -  2*  •  33  •  5  .  11  a?tf  +  5  .  38  •  11  a8ar« 
-  23  -  37  • 


It  is  not  necessary  or  desirable  to  perform  the  multiplica- 
tion in  such  an  expression  as  22  •  33  •  5  -  11  ;  such  terms,  if 
desired  numerically,  are  usually  obtained  progressively  from 
preceding  terms  as  in  example  (e)  above. 


PROBLEMS 


1.   Expand     to     6     terms,    (a  +  x)6,    (a  +  a)14,    (a  +  2  x)10' 
(a  -  3  x)9. 


BINOMIAL  SERIES  AND  APPLICATIONS         197 

2.  Write   5   terms   in   simplest   form    (prime   factors)   of 
(1  +  *)6,  (1  +  a)14,  (1  +  2  x-)10,  (1-3  a-)9. 

3.  Compute    to    4    decimal    places    (1  -f  .05)6,    (1  -f-  .05)", 
(1  +  -05)10,  (1  -  .05)9. 

4.  Compute  to  2  decimal  places  the  value  at  the  end  of 
10  years  of  $  100  placed  at  interest  at  6  %  compounded  annu- 
ally ;   use   100  X  (1  +  .06)10.      How   could  you  use  the  result 
obtained  to  find  the  value  at  the  end  of  20  years  ? 

5.  From  problem  3  give  the  amount  at  the  end  of  6,  14, 
and   10   years,  respectively,  of    $  256  at  5  %    interest,  com- 
pounded annually. 

6.  Find  the  value  at  the  end  of  6,  14,  and  10  years  respec- 
tively of  an  annuity  of  1  per  annum,  paid  at  the  end  of  each 

year,  interest  at  5  %.     Use  the  formula  s^  =  ^ — -*-—* and 

the  preceding  results. 

7.  Compute  the  values  in  3,  4,  and  5  by  logarithms  and 
compare. 

8.  Given  210  =  1024,  find  to  1   decimal   place    (2  +  .Ol)10. 
Ans.   1077.7. 

Find  also  (2.1)10  to  one  decimal  place,  and  check  by  logs. 

9.  Find  (12.3)3  to  five  significant  figures. 

10.  Find  the  amount  at  the  end  of  20  years  of  $  100  placed 
at  interest,  3%,  compounded  semiannually. 

11.  Write  the  8th  term  of  (1  -  3x)"  and  of  (1  -f  a?)». 

12.  How   many   terms    in    (1—3 a;)17?     Write   the    middle 
terms. 

13.  Write  in  simplest  form  the  coefficient  of  &  in  (1  —  2  #)28. 

14.  Write   the    series    for    (1  +•  a;)5,    (1  -f-  a;)6,    (1  -f-  a;)7,   and 
(1  +  x)s.     What  is  the  sum  of  the  coefficients  ? 

(NOTK.  —  Substitute  1  for  x.) 

15.  What  is  the  sum  of  the  coefficients  in  (1  +  x)19? 

16.  Write  7  terms  of  (1  +Va;)10  and  of  (1 


198  UNIFIED  MATHEMATICS 


2.    Proof   of   (a  +  *)"  =  a"  4-    a"~1jr  +       ~       a"-2*2  + 

1  1  •  2 

mi  TKRSI  TO  A  TOTAL  OF  (r  —  1)  V\>  -ioi:s 

n(n-l)(n-2)       ,  .  n(n-l)(n  -  2)  -          r+1 

1.2-3  1-234      (r-1) 

The  proof  of  this  expansion  for  positive  integral  values  of  n 
is  effected  by  the  process  called  mathematical  induction. 

Evidently  (a  +  x)2  =  a2  +  2  ax  +  xz,  follows  the  rule  ; 
also  (a  +  a)3  =  a3  +  3  a?x  +  3  ax2  +  Xs,  follows  the  rule. 

By  the  rule, 

,4,      ,4-3,      ,4.3-2         ,4.3-2.1, 

(«  -f  a;)4  =  a^  +  _  a3a.  +  __  a2a.2  +  __  __  aa,3  +  —j-^-—  X* 


=  a4  +  4  a3«  +  6a2a;2  +  4  aic3  +  a;4  ; 

by  actual  multiplication  we  find  the  same  series,  showing  that 
the  rule  as  given  holds  for  n  =  4. 


1.2-3-(r-2) 

mi  TERM 

4.  n(n-l)(n-2)-.(w-r  +  2)     _r+1     ,  . 
1  -  2  -  3  -  (r  -  1) 

multiply  by  a  +  x  =  a  +  x 

(a  +  *)"+1  =  an 


THE  NEW  mi  TERM 

n(n  -  l)(?i  -  2) ...  (n  -  r  +  3)(n  -  r  +  2) 
1.2-3-  (r  — 2)(r  — 1) 

n(M  _  i)(n  _  2) ...  (n  -  r  +  3)\    -^j^ 


BINOMIAL  SERIES  AND  APPLICATIONS         199 

Note  that  these  two  coefficients  of  the  new  rth  term  have 
the  first  (r  —  2)  factors  of  numerator  and  denominator  the 
same ;  multiply  the  denominator  and  numerator  of  the  second 
term  by  r  —  1  and  then  add  the  numerators,  taking  out  the 
common  factors,  giving 

NEW  mi  TERM  COMMON  REMAINING  REMAINING 

FACTORS  FACTOR  or  FACTOR  OP 

FIRST  TERM  SECOND  TERM 


I  1.2.3.4.5...  (r-1) 

The  new  rth  term  may  be  written,  then, 

(n  +  l)(n)(n  -  l)(n  -  2)  ...  (n  -  r  +  3)     (I1+1)_r+1      x . 
1.2- 3. 4».  (r-1) 

the  rth  term  of  (a  +  #)B+1  is  formed  according  to  the  rule  with 
(>i  +  1)  substituted  throughout  for  n ;  the  numerator  contains 
(r  —  1)  factors  beginning  with  (n  -\-  1),  and  the  denominator 
contains  the  same  number  of  factors  beginning  with  1. 

Hence  if  this  expansion  assumed  for  (a  +  a;)"  is  correct  for 
any  value  n,  it  is  correct  for  a  value  one  greater,  n  +  1.  The 
theorem  is  true,  by  trial,  for  n  =  4  ;  hence  it  is  true  for  n  =  5  ; 
since  it  is  true  for  n  =  5  it  is  also  true  for  n  =  6 ;  ...  and  so 
for  every  integral  value  of  n. 

3.    Binomial  series ;  any  exponent.  —  The  equation, 

•  n(n-l)(n.-2)(n-8)    4 
"1.2.3.4 

can  be  shown  by  methods  of  the  higher  mathematics  to  hold 
for  all  values  of  n  when  —  1  <  x  <  1.  This  means  that  a 
series  for  (1  +  x)*  can  be  obtained  by  substituting  n  —  ^  in 
the  above  formula.  Similarly  (l+x)~3  and  (1  +  #)~7,  (1  +  cc)~^ 
or  (1  -}-  »)V?  can  be  developed  in  powers  of  x,  when  |a;|  <  1,  by 


200  UNIFIED  MATHEMATICS 

substituting  for  n  the  values  —  3,  and  —  7,  and  the  like  in  the 
above  formula.  These  series  are  of  frequent  use  in  statistical 
work.  Thus  if  money  is  worth  6%  per  annum  the  interest 
for  one  half  year  is  not  3  %  of  1,  since  this  rate  continued  for 
the  full  year  would  give  at  the  end  of  one  year  in  addition  to 
the  6  °Jo  of  1,  the  interest  on  the  3  %  of  1  for  one  half  year ; 
the  interest  on  1  for  one  half  year  is  taken  to  be 

j=(l  +  . 06)* -1, 

and  this  rate  of  interest  per  half  year  accumulates  at  the  end 
of  two  half  years  a  principal  of  one  to  (1.06),  or  is  equivalent 
to  6  %  per  year.  Similarly  the  effective  rate  of  6  %  per 
annum  means  that  the  interest  for  one  fourth  of  a  year  will 
not  be  .015  times  the  principal,  but  rather  (1  +  .06)*  —  1, 
since  this  is  the  rate  of  interest  for  one  quarter  of  a  year 
which  continued  for  four  quarters  will  accumulate  a  principal 
of  1  to  1.06,  since  [(1  +  .06)*]*=  1.06. 

In  our  illustrative  problems  we  will  assume  that  the 
terms  that  follow  any  given  term  in  the  expansion  of  an  ex- 
pression like  (1  +  x)*  are  together  less  than  the  last  term 
given ;  the  general  proof  of  the  convergence  of  these  series 
is  reserved  for  the  calculus ;  however,  it  is  evident  here  that 
each  new  factor,  when  x  is  less  than  1,  diminishes  in  value  and 
finally  the  terms  are  in  turn  respectively  less  than  the  terms 
of  a  geometrical  series  with  ratio  x ;  thus,  below  in  the  ex- 
pansion of  (1  -}-  .06)*,  beyond  any  given  term  the  terms  are 
respectively  less,  term  by  term,  than  the  terms  of  a  geometri- 
cal series  with  ratio  .06 ;  the  sum  of  all  terms  beyond  the 

fourth  term  is  certainly  less  than  — ,  wherein  a  is  the  fourth 

.94 

term,  since  the  corresponding  geometrical  series  even  to  an 
infinite  number  of  terms  has  only  this  sum,  -  —  =  —  • 

Note  particularly  that  the  expansion  of  (1  +  x)n  for  values 
of  n  other  than  positive  integers  leads  to  a  series  which  has 
no  termination,  i.e.  to  an  infinite  series ;  this  series  is  valid 


BINOMIAL  SERIES  AND  APPLICATIONS        201 

and  has  meaning  only  when  |  x  \  <  1  ;  similarly  any  infinite 
series  given  by  (a  +  x)n  has  meaning  only  when  --  <  1.  In 

our  further  discussion  this  limitation  will  be  consistently 
assumed. 

As  an  illustration  of  the  possible  absurdity  from  the  point 
of  view  of  finite  series  of  the  infinite  series  given  by  (1  +  x)n, 

let  us  take  the  fraction  -      -  which  may  be  written  (1  —  a?)"1, 

and  developed  by  the  binomial  theorem  as, 

l  +  a3  +  #2  +  ar5  +  :E4  +  2ri-|-  .... 

For  values  of  x  numerically  less  than  unity  this  series  is  valid, 
but  if  you  put  x  equal  to  3,  or  —  5,  or  other  value  greater  than 
unity,  you  obtain  an  absurdity. 

4.   Illustrative  problems.  —  a.   Compute  (1  +  .06)^  to  5  deci- 
mal places. 


(1  +  .06)*  =  1  +  i(.06)  -  1  (.06)2  +  ^  (.06)3  +  ... 
=  1  +  .03  -  .00045  +  .0000135  +  ... 
=  1.02956. 

If  the  value  of  (1.0ft)*  were  desired  to  eight  decimal  places,  progressive 
computation  would  be  desirable,  appearing  as  follows  : 

Ti  =  1  4)  .0000135 

T2  =  i(.OO)  TI  =  .03  .000003375 

=  -.  00045 


2  \  2  /    '  .000016875 


;  =  +.  0000135 


-03 


2  3  .00000050625 

.042 
1012 
2024 


.00000002125 

T5  =  -  ^ .  'I*  T4  =  -  .000000506  1  -030013521 

-  .000450507 
T6  =  _  I .  M  T&  =  +  .000000021  1.029563014 

r7  =  -? . ^  T6  =-  .000000001 

w    0 


202  UNIFIED  MATHEMATICS 

The  letters  TI,  T2,  T3,  T4,  •••  represent  the  first,  second,  third,  fourth, 
and  succeeding  terms  ;  each  term  is  obtained  from  the  preceding,  intro- 
ducing the  new  factors. 

6.   "Write  six  terras  of  the  series  for  (1  +  x)^  and  for  (1  —  as)*. 
1    -  1  3          i  -i    -3   -5 


1    riJ.^S.jzJi    -7 
2'    2    '    2    '    2    '    2 

1.2.3-4.5        - 
=  l  +  |x-  Jx-  r2-lx.  T,-Jx.  T4-^x-  T,  ----  . 
Alternate  terms  after  the  second  are  positive  and  negative. 

I.  =1.^3          1-1-3     -5 
2  '    2    '    2  2  '    2    '    2    "    2 


1/-J\ 
1          2\  2   / 


1.2.3.4 
1-1-3      -5      -7 

2  '     2     '     2     '     2          2 


1.2.8.4.6 


=  l-ix  +  Jx- 
Every  term  after  the  first  is  negative. 

c.   Compute  V.98  to  5  places. 


-  .02)    =  1  -  J(-02)  +      ~       (.02)2  _      ~-       (.02)3  +  ... 
1*3  1*8*3 

=  l  _  .01  -  .00005  -  .0000005  -  — 
=  .9899495  or  .98995  to  5  places. 

d.   Compute  the  cube  root  of  (1012)  to  6  significant  figures. 
(1012)5  =(1000)3(1  +  .012)^  =  10(1  +  .012)i 


(1  +  .012)*  =  1  +  K-012)  +  *(\  ~  ^  (.012)2  +  Ki  -  !)(^  ~  2)  (.012)3  +  ... 
1  •  2t  1  •  £  •  o 

=  1  +  .004  -  .000016  +  .0000001  —  ... 
=  1.003984. 
(1012)'  =  10  x  1.003984  =  10.03984. 

e.   Compute  (1.05)5  to  4  places  ;  treat  this  as  1.05  x  (1.05)^  ; 
.05  may  be  taken  as  ^.     Check  by  logarithms. 


BINOMIAL  SERIES  AND  APPLICATIONS 


203 


/.   Compute  the  square  roots  of  26  and  30  to  4  places. 

V26  =  26^  =  (25  +  1)2  =  5(1  +  '&)$  =  5(1  +  .04)* 

=  5(1  +  .02  -  .0002  +  .000004) 
=  5(1.019804)  =5.09902. 
30^  =(25  +  5)2  =  5(1  +  .2)*  =  6(1  +  .1  -  .005  +  .0005  -  .0000625) 

=  5(1.0954375)=  5.47718. 
Check  roughly,  using  logarithms. 

g.   Compute  (1.05)~7  to  5  significant  figures. 
(1.05)-*  =  1  -  7(.05)+-7(-8)(.06)«  +  ~7(1~^(3~9)(.05)3 


I  ----  Y      _----r      y, 

1-2-3-4  1-2-3.4-5 

=  l  _  .35  +  .07  -  .0105  +  .0013125  -  .0001444  +  .0000144  =  .71068. 

5.  Historical  note.  —  The  binomial  theorem  as  applied  to 
(a  +  b)2  and  (a  +  6)3  was  well  known  to  Euclid  (320  B.C.)  and 
other  early  Greek  mathematicians.  The  great  Arabic  mathe- 
matician and  poet  Omar  al  Khayyam  (died  1123  A.D.)  extended 
the  rule  to  other  positive  integers.  In  China  there  appeared 
in  1303  a  work  containing  the  binomial  coefficients  arranged 
in  triangular  form,  the  so-called  Pascal  (1623-1662)  triangle 

of  coefficients. 

11111111 
1234567 
13    6  10  15  21 
1  4  10  20  35 
1  5  15  35 
6  21 


1 

1     1 

121 

1331 

14641 

15    10  10   51 

1  6  15  20  15  6  1 


1 

1  7 
1 


PASCAL'S  TRIANGLE  CHINESE  FORM  OF  TRI- 
OF  COEFFICIENTS,  ANGLE  OF  COEFFI- 
PRINTED  1665.  CIENTS. 

m 


121 
1331 
14641 
1  5  10  10    51 
1  6  15  20  15  6  1 

STIFEL'S  (1486-1567) 
TRIANGLE  OF  COEFFI- 
CIENTS, 1544. 


The  general  rule  for  any  exponent  --  was  first  discovered 

n 

by  Sir  Isaac  Newton,  and  made  known  in  a  letter  of  date  Oct. 


204  UNIFIED  MATHEMATICS 

24,  1676,  to  a  friend  named  Oldenburg.  It  is  of  interest  to 
note  that  Newton  wrote  each  coefficient  in  terras  of  the  coeffi- 
cient immediately  preceding,  following  the  lines  indicated  in 
our  numerical  problems  above.  The  complete  proof  for  the 
general  case,  any  real  or  complex  imaginary  number,  was  finally 
effected  by  a  brilliant  Norwegian  mathematician,  Abel  (1802- 
1829),  in  1826. 

PROBLEMS 

1.  Find  the  coefficients  of  the  first  five  terms  of  (1  +  x)* 
and  (1  —  x)*  and  use  them  in  the  next  two  problems. 

2.  Compute  to  3  significant  figures  the  square  roots  of  27, 
28,  and  29  as  5(1  +  .08)*,  5(1  +  .12)*,  and  5(1  +  .16)*,  respec- 
tively. 

3.  Compute  the  square  roots  of  the  first  eleven  integers, 
to  5  places,  taking  \/2  as  ^(196  +  4)*  —  |£(i  +  ^i,  V3  as 
2(1  -£)*,    V5   as  2(_l  +  i)*,_V6_as    (4+  2)1  =  2(1  +  .5)*, 
V7  as  1(25  +  3)*,  V8  as  2V2,  VH  as  £(100  -  1)*.     Check 
the  first  four  significant  figures  by  logarithms. 

4.  Write  5  terms  of  (1  -+-  »)«,  and  of  (1  —  #)*. 

5.  ITse  these  to  compute  the  values  of  A/7  and    ^9,  as 


6.  From  the  cube  root  of  9  find  the  cube  root  of  3. 

7.  Find  the  cube  root  of  6. 

8.  Using  the  cube  roots  of  6  and  3,  find  the  cube  root  of  2. 

9.  Compute  (1  +  .05)*  to  5  places. 

10.  Write  5  terms  of  (2  -  3  *)~*. 

11.  Write  the  6th  term  of  (2  - 


12.    Find  the  value  to  4  significant  figures  of 


expanding  (1  -  .02)~*.  ^  ~  -02 


BINOMIAL  SERIES  AND  APPLICATIONS         205 
13.   Expand  in  powers  of  i  to  4  terms  -  and  -  •    What 


is   the  approximate    percentage   error   in   using   (1  —  i)   and 
(1  -f  i)  as  multipliers,  respectively,  instead  of  -  :  and  -   —  ., 

J-  *T~  1  JL  ~~*  1 

when  i  =  .01,  .05,  .005,  and  .5,  respectively  ? 

14.  Find  the  fifth  root  of  35  correct  to  2  decimal  places. 
What  is  the  shortest  way  ?      Compute  the  root  to  5  places. 
What  method  can  you  use  ? 

15.  Write    the    term    containing    x6   in   the    expansion    of 
(1-3*)-*. 

16.  Write  the  first  five  terms  of  (10  +  .3)8;    of  (10.3)6;  of 
(10.3)~6  and  give  the  value  to  4  significant  figures. 

17.  Time  yourself  on  writing  and  simplifying  10  terms  of 
the  following  five  expansions  :  (1  +  x)  *,  (1  —  x)'5,  (1  —  2  »)"*, 


18.    Compute   correctly  to  4  significant   figures,  using   the 

formulas  for  -  and  --     Time  yourself  on  the  exercise. 
1+i          1  -  i 

18      27       54      62      68 


.98'  1.02'  1.03'  .99'  1.05 

19.    Compute  the  following  to  three  significant  figures,  tim- 
ing yourself : 

(1.06)*,  (1.06)T2,  (1.05)*,  (1.06)*,  (1.10)*,  (1.06)-*. 


CHAPTER   XIII 


RIGHT    TRIANGLES 

1.  Right  triangles.  —  To  apply  our  trigonometric  work  to  the 
numerical  solution  of  right  triangles  place  the  triangle  under 
consideration  in  quadrant  I  in  proper  position  to  be  able  to  read 
the  trigonometric  functions  of  one  acute  angle. 

x  =  rcos  a  =  rsin  p 
y  =  r  sin  a  =  r  cos  p 

tan  a  =  -  =  cot  p 


--c 


Fundamental  formulas  of  the  right 
triangle 


cot  a  =  -  =  tan  B 
y 


a  +  B  =  90°. 
The   equation   x  =  r  cos  a  may    be    written 


cos 


x 
a  =  - 

r 


or 


cos  a 


f  as  occasion  demands  ;  similar  transformations  are  to 


be  effected  upon  the  other  equations  given. 

Given  a  with  x,  y,  or  r;  or  (3  with  x,  y,  or  ?•;  or  two  of  the 
lengths ;  these  formulas  enable  us  to  solve  the  right  triangle 
completely  for  the  remaining  three  parts. 

The  student  is  advised  to  draw  the  figure  to  scale  on  coordi- 
nate paper,  using  a  protractor  to  lay  off  correctly  to  degrees  the 
angles  given,  before  attempting  to  apply  any  formulas ;  then 
write  the  required  equations  directly  from  a  consideration  of 
the  figure,  and  not  by  attempting  to  memorize  t;he  solutions 
for  the  different  types  of  problems.  The  lengths  as  given 
graphically  serve  as  a  check  upon  the  values  obtained. 

206 


RIGHT  TRIANGLES 


207 


2.   Right  triangles.     TYPE  I.  —  Given  hypotenuse  and  one 
angle,  i.e.  a  and  r,  or  /3  and  r. 

A  guy  wire  168  feet  long  reaches  to  the  top  of  a  tall  chimney, 
making  an  angle  of  37°  with  the  ground.  Find  the  height  of 
the  chimney  and  the  distance 
from  the  supporting  peg  to 
the  foot  of  the  chimney. 


\J! 

/  i 

-  ^ 

__5  

i    ft?      :"" 

Gt 

,/fl 

+  ' 

slid 

X  «,•  a      OT-     - 

"jffS82    I 

Jf   * 

-1r=m  

~\ 

m 

so^po-ifto-iji^q 

sills 

Solution.  —  First  draw  the  fig- 
ure, as  indicated,  using  ^  inch  to 
represent  30  units. 

y  -  168  sin  37° 
x  =  168  cos  37° 
Check,     y  =  x  tan  37°  Hypotenuse  and  one  angle  given 

Using  natural  functions  there  are  here  three  problems  in  multiplication. 
The  logarithmic  solution  is  as  follows  : 

log  168  =      2.2253  log  168  =      2.2253 

log  sin  37°  =      9.7795-10        log  cos  37°  =      9.9023  -  10 
logj/  = 


logx=      2.1276 
x  =  134.2. 


2.0048 
y  =  101.1. 

Check,     log  x  =  2.1276 
log  tan  37°  =  9.8771  -  10 
log  y  =  2.0047. 

Compare  with  preceding  value  of  log  y  ;  a  dis- 
agreement in  the  fourth  place  is  permissible.  It 
would  not  affect  the  fourth  significant  figure  of 
y  in  this  case  ;  nor  would  the  measurements  of 
height  of  chimney  and  length  of  guy  wire  be 
made  with  greater  accuracy  than  to  one  tenth  of 
a  foot,  x2  +  y2  —  1682  could  be  used  as  a  check. 


.Ill 


Angle  and  side  given 


3.  Right  triangles.  TYPE  II.  —  Given 
one  leg  and  an  angle. 

If  a  telegraph  pole  is  34  feet  high,  and 
the  supporting  wire  makes  an  angle  of  62° 
with  the  ground,  find  the  length  of  the 
wire  and  the  distance  from  the  foot  of  the 
supporting  peg  to  the  foot  of  the  pole. 


208 


UNIFIED  MATHEMATICS 


The  corresponding  formulas  are  as  follows  : 


x  =  34  cot  62° 
34 


Check,     x-r  cos  62° 
or  342  =  r2  -  x2 


sin  62° 

log  34=    1.5315 
log  cot  62°  =    9.7257  -  10 
logx  =    1.2572 
x  =  18.08. 


log  34  =  11.6315-  10 

-  log  sin  62°  =    9.9459  -  10 
logr=    1.5856 

r  =  38.51. 
Check,     logr  =  1.5856 
+  log  cos  62°  =  9.6716  —  10 

logx  =  1.2572,  checks. 

4.    Right  triangles.      TYPE  III.  —  Given  a  leg  and  the  hy- 
potenuse. 

Let  the  side  a  =  341  and  c,  the  hypotenuse,  equal  725. 
Evidently,    sin  a  =  f  f^, 


b  =  725  cos  a. 

log  341  =  12.5328  -  10 
log  725  =    2.8603 
log  sin  «  =    9.6725  —  10 
«  =  28°  4'. 


Check,     b  =  341  cot  a. 


log  725  = 
log  cos  a  = 


Hypotenuse  and  one  side  given 

Graphical  solution  gives  a  rough  check 

to  two  significant  figures. 


2.8603 
9.9457  -  10 
=      2.8060 
6  =  639.7. 

Check,     log  341  =  2.5328 

log  cot  a  =    .2731 

log  b  =  2.8059. 

Compare  with  above  value 
log  b  =2.8060  as  a  check; 
the  error  of  1  here  is  in- 
evitable with  4-place  log- 
arithms. 

Note  that  on  the  small 
graph  only  two  places  are 
accurately  representable. 


5.    Right  triangles.     TYPE  IV.  —  Given  the  two  legs. 

Entirely  similar  except  that  the  initial  formula  is  for  tan  a  instead  of 
sin  «. 

Note  that  commonly  in  lettering  right  triangles  x  and  y  or  a  and  b  are 


RIGHT  TRIANGLES  209 

used  for  the  legs,  r  or  c  for  the  hypotenuse,  a  and  p  for  the  angles  at  A  and 
.B,  opposite  a  and  b  respectively. 

In  Type  III  if  a  and  c  are  nearly  equal  we  may  avoid  the  use  of  the 
sine  of  the  angle  near  to  90°  by  computing  the  other  side  using  the 
formula 

&2  =  c2  —  a2  =  (c  —  «)(c  +  a). 
Thus  if          a  =  718,  c  =  725,  62  =  (725  -  718)(725  +  718) 

=  7(1443)  =10101. 
b  =  100.5, 

either  by  inspection  as  in  this  case,  or  from  a  table  of  squares,  or  by 
logarithms. 

PROBLEMS 

Solve  the  following  right  triangles  by  logarithms : 

1.  Given  r  =  240,  «  =  30°  10'. 

2.  Given  a  =  368,  a  =  30°  14'. 

3.  Given  a  =  368,  r  =  579. 

4.  Given  a  =  368,  6  =  275. 

5.  Solve  for  the  missing  parts  the  following  ten  problems, 
using  logarithms ;  time  yourself ;  the  exercise  should  be  com- 
pleted within  30  minutes. 

a.  Given  r  =  186,  a  =  84.3.  e.  Given  a  =  930,  a  =  24°. 

b.  Given  a  =  .394,  b  =  .654.  /.  Given  6  ==  184,  a  =55°  15'. 

c.  Given  a  =•  2.89,  £  =  68°  24'.  g.  Given  r  =  .0936,  b  =  .0418. 

d.  Given  b  =  706,  a  =  70°  10'.  h.  Given  6  =  3.24,  ft  =  86°  14'. 

i.    Given  b  =  878,  a  =  48°  19'. 

j.    Given  r  =  8.4  x  106,  ft  =  34°  16'. 

6.  Area.  —  In  computations  of  functions  involving  measured 
and  computed  values,  measured  values  are   taken,  as   far  as 
possible,  in  preference   to  computed  values.     The   computed 
value  involves  not  only  the  inaccuracies  or  errors  of  measure- 
ment, but  also  the  errors  of  computation,  the  inevitable  errors 
of  computation  with  approximate   numbers  as  well   as   the 
avoidable  errors.     Among  the  following  formulas  for  the  area 
of  a  right  triangle  the  student  should  select,  in  accordance  with 


210 


UNIFIED  MATHEMATICS 


the  principle  mentioned,  the  formula  to  be  used  in  each  prob- 
lem involving  a  right  triangle. 

A  =  %  ab  =  \  a2  cot  a  =  \  62  tan  a  =  \  c2  sin  a  cos  a. 


7.    Applications.  —  In   the    application    of    the    solution   of 
right  triangles  to  practical  problems  we  find  that  the  difficulty 

is  frequently  a  matter  of  ter- 
minology rather  than  of  prin- 
ciple. The  student  is  urged 
to  acquire  some  real  famili- 
arity with  the  industrial  and 
scientific  application  of  the 


-rA 


1? 


Common  terms  relating  to  angles 

Dip — depression — elevation — bias 
—  departure. 


principles  explained. 

The      terms      "  elevation," 

"  depression,"  "  dip,"  "  de- 
parture," and  "  bearing,"  all  refer  to  angles.  Thus  in  the  figure 
ABC,  if  A  C  is  in  the  direction  of  the  sun  or  if  C  represents  the 
top  of  a  mountain  viewed  from  A,  then  angle  BAG  is  termed  the 
angle  of  "  elevation  "  of  C  as  viewed  from  A :  if  the  observer 
is  at  C,  on  a  mountain  or  in  an  airship,  HCA  is  the  angle  of 
"  depression  "  ;  if  A  represents  the  horizon  as  viewed  from  C, 
then  HCA  is  called  the  "  dip  "  of  the  horizon.  If  CA  repre- 
sents a  vertical  section  of  a  vein  of  coal,  the  angle  HCA  or 
BAG  is  called  the  "  dip "  of  the  vein ;  in  navigation  if  AB 
represents  east,  then  angle  BAG  represents  the  "  departure  " 
north  of  the  line  AC,  whereas 
in  surveying  the  angular  de- 
flection from  north  or  south  is 
given  as  the  "  bearing." 

Frequently  some  function 
of  an  angle  is  given  from 
which  the  angle  must  be  de- 
termined. The  pitch  of  a  roof 
is  given  as  the  height  divided 
by  the  span,  whence  the  corresponding  slope  angle  of  the  roof 
is  6,  given  by 


Pitch  =  - 
2s 


RIGHT  TRIANGLES 


211 


tan  0  =  -, 
s 

where  h  is  the  height  and  2  s  is  the  total  span. 

The  slope  of  a  railroad  is  commonly  given  as  so  many  (K) 
feet  of  rise  in  100  feet  horizontally ;  this  gives  the  slope  angle 

0  from  tan  6  =  —  =  h  %. 
1UO 

A  spiral  thread  winds  about  a  cylinder  advancing  a  height 
h,  called  the  "  lead,"  in  one  complete  turn ;  the  circumference 


, .  I , 


Full  size  representation  of  a  one-inch  cylindrical  screw 
The  "  lead  "  is  2%  of  an  inch. 

of  the  cylinder  is  the  base  and  the  "lead"  is  the  altitude  of 
a  right  triangle  which  may  be  regarded  as  wrapt  about  the 
cylinder  to  give  the  spiral.  The  angle  a  made  with  h  by  the 
spiral  line  is  called  the  angle  of  the  spiral ;  evidently 

tan  a  =  -^-  • 
h 


Circumference,  AC,  and  length  of  one  spiral,  AB,  of  the  above  one-inch 
cylindrical  screw 


212 


UNIFIED  MATHEMATICS 


PROBLEMS 

1.  A  standpipe  subtends  an  angle  of  4°  at  the  eye  of  an 
observer ;  if  its  height  is  280  feet  above  the  level  of  the  eye, 
find  its  distance  from  the  observer. 

2.  If  the  diameter  of  the  standpipe  in  1  subtends  an  angle 
of  15'  at  the  eye,  what  is  the  diameter  in  feet  ?     Suppose  that 
the  angle  at  the  eye  lies  between  10'  and  20',  what  range  of 
diameter  would  these  values  give  ? 

3.  The  shadow  of  a  flagstaff  60  feet  high  is  48  feet  long. 
Find  the  angular  elevation  of  the  sun. 

4.  Using    trigonometric    functions,  find    the    height   of  a 
building  which  at  the  same  time  casts  a  shadow  87  feet  long. 

5.  Find  the  lengths  of  the  circle  of  latitude  and  the  circle 

of  longitude  through  your  home 
city. 

6.  When  the  sun  is  directly 
over  the  equator,  the  latitude  of 
any  place  on  the  earth's  surface 
from  which  the  sun  is  visible  is 
the  angle  between  the  zenith  line 
(the  vertical)  and  the  line  to  the 
sun,  when  the  sun  is  on  the  me- 
ridian. Find  the  shortest  length 
of  the  shadow  of  a  pole  100  feet 
high,  at  noon,  latitude  40°  N.,  and 
also  for  latitude  42°  18'  N. 


Q 


Zenith    distance    representing 

latitude 

Sun  on  celestial  equator,  di- 
rection OE.  OP  is  zenith 
direction  of  P. 


7.  Compute  the  diameter  of  a  circle  circumscribed  about 
an  equilateral  triangle  of  side  40. 

8.  Find  in  a  circle  of  radius  486  cm.  the  chords  of  angles 
of  30°,  60°,  45°,  90°,  120°,  72°,  68°. 

9.  For  any  angle  a,  find  the  chord  and  the  chord  of  half 
the  angle  in  terms  of  the  radius  r.     Apply  the  latter  formula 
to  obtain  the  results  of  problem  8. 


RIGHT  TRIANGLES  213 

10.  A  pendulum  of  length  34  inches  swings  between  two 
points  10  inches  apart ;  compute  the  arc  of  the  swing.     If  this 
is  a  seconds  pendulum,  passing  the  vertical  once  every  sec- 
ond, what  is  the  velocity  of  the  pendulum  bob?     Find  the 
chord  of  this  arc,  and  the  difference  between  the  chord  and  the 
arc. 

11.  A  circular  arch  over  a  doorway  is  to  be  4  feet  wide  and 
20  inches  high ;  compute  the  radius.     Compute  for  heights 
of  10  to  24  inches  by  2-inch  intervals. 

12.  Given  the  radius   10   feet  and  the   span  4  feet  of  a 
circular  arch.     Compute  the  height.     Compute  for  spans  of 
2  feet  to  20  feet,  by  2-foot  intervals. 

13.  Adapt  the  preceding  results  to  a  radius  of  8  feet.     How 
closely  would  interpolation  give  correct  results  ?    discuss  by 
considering  the  problem  graphically, 

14.  In  a  circle  of  radius  100  inches,  compute  to  one  decimal 
place  the  lengths  of  sides  and  the  perimeters  of  regular  in- 
scribed polygons  of  3, 4,  5,  6,  7,  8,  9, 10, 11,  and  12  sides.     Time 
yourself  on  the  exercise.     State  the  general  formulas  involved. 

15.  Compute  perimeters  of  regular  circumscribed  polygons 
of  3,  4,  5,  6,  7,  8,  9,  10,  11,  and  12  sides  in  a  circle  of  radius 
100.     Time  yourself  on  the  numerical  work ;  30  minutes  is 
ample  time. 

16.  Compute  the  perimeter  of  a  regular  inscribed  polygon 
of  96  sides,  and  of  a  regular  circumscribed  polygon   of  96 
sides,  radius  100.     How  does  the  circumference  compare  with 
these  two  values  ?     Archimedes  computed  these  lengths  by 
plane  geometry  methods  and  so  found  ?r  to  lie  between  3|  and 
3|^.     Check  his  result. 

17.  Frequently   arches  of  bridges  are  circular    segments; 
find  the  radius  of  the  circular  arch  of  the  famous  Rialto  in 
Venice  (see  illustration,  page  225).     The  width  of  the  arch  is 
95  feet  and  the  height  is  25  feet.     Draw  the  graph  of  the  arch 
to  scale. 


214 


UNIFIED  MATHEMATICS 


18.  One  of  the  largest  masonry  bridges  in  the  U.  S.  is  the 
Rocky  River  bridge  at  Cleveland ;  the  height  of  the  circular 
arch  is  80  feet  and  the  span  is  280  feet.     Find  the  correspond- 
ing radius. 

19.  Find  the  angle  of  the  spiral  represented  in  the  above 
diagram  of  the  one-inch  cylindrical  screw. 

20.  Find  the  angle  of  a  cylindrical  screw  of  diameter  \  inch 
which  advances  ^  of  an  inch  in  one  complete  turn. 

21.  A  twelve-inch  gun  has  a  muzzle  velocity  of  approxi- 
mately 2500  feet  per  second  (f.s.).     The  velocity  is  tested  by 


Determination  of  velocity  of  a  projectile 
One  screen  is  at  the  muzzle  of  the  gun. 

electrical  means  ;  screens  are  placed  at  a  known  distance  apart 
and  the  projectile  in  passing  through  the  screens  breaks 
successively  two  electrical  circuits  which  serve  to  give  the 
time  of  flight  of  the  projectile  to  thousandths  of  a  second 
in  passing  through  the  known  distance.  The  apparatus  may 
also  be  used  to  determine  the  angle  of  elevation  of  the  larger 
guns.  In  the  figure  TAT  represents  the  axis  of  trunnions  of 
a  twelve-inch  gun  ;  AM  along  the  axis  of  the  barrel  is  25  feet ; 
MS  is  180  feet ;  the  one  screen  is  over  the  muzzle  and  the 
other  screen  is  at  a  height  of  94  feet  above  the  axis  of  trun- 
nions. Determine  the  angle  of  elevation  of  the  gun  and 
reduce  to  "  mils."  Find  the  horizontal  distance  MQ  between 


RIGHT  TRIANGLES 


215 


muzzle  and  screen  and  the  vertical  distance  between  muzzle 
and  screen,  QS ;  find  the  time  of  flight  of  the  projectile, 
assuming  2500  f.s.  as  velocity ;  find  "  horizontal  velocity,"  vt, 
and  "  vertical  velocity,"  vv,  by  dividing  M Q  and  QS  re- 
spectively by  this  time  of  flight.  The  vertical  velocity  vv 
divided  by  32.2  gives  approximately  the  time  in  seconds  that 
the  projectile  will  continue  to  rise  ;  find  this  time ;  the  position 
of  the  projectile  after  this  interval  of  time  is  given  approxi- 
mately by  the  product  of  horizontal  velocity,  vx,  multiplied  by 
the  time,  as  horizontal  distance  from  the  gun,  and  by  vertical 
component  of  velocity  multiplied  by  the  same  value  of  t  less 
16.1  multiplied  by  t2,  as  ordinate.  The  equations  are 

x  =  vrt, 

y  =  vvt-  16.1Z2. 

What  error  is  possible  in  the  angle  measured  if  the  height  of  S 
is  given  only  within  one  foot  ?  The  aim  is  directed  at  a  point 
two  feet  below  the  top  of  the  screen,  as,  in  general,  there  is  a 
slight  "  jump "  due  apparently  to  the  explosion.  Estimate 
the  jump  in  degrees  and  minutes,  and  in  "  mils  "  if  the  pro- 
jectile hits  the  top  of  the  screen. 

22.   When  two  screens  are  used  with  a  large  gun  the  dis- 
tance between  screens  is  sometimes  measured  by  taking  equal 


Distances  of  screens  from  muzzle,  M,  determined  by  right  triangles 
Two  screens  spaced  100  feet  apart  (horizontally). 


distances  MH  and  ME  at  right  angles  to  the  line  MS^Sz  and 
measuring  the  angles  MHS^  MHS2,  MESi,  and  MES*  Note 
that  the  screens  are  20  to  100  feet  in  the  air  on  tall  standards, 


216 


UNIFIED  MATHEMATICS 


making  it  inconvenient  to  measure  the  distance  with  a  steel 
tape.  Assuming  that  the  distances  ME  and  MH  are  taken  as 
20  feet  and  that  the  angles  MHSl  =  MES^  =  70°  10',  and  that 
angle  MHSZ  —  MES2  =  82°  34',  compute  the  distance  MSlt 
MSz,  and  S^S*.  If  the  screen  Si  is  at  an  elevation  of  34  feet 
and  the  screen  S2  is  at  an  elevation  of  83  feet,  compute  the 
angle  of  elevation  of  the  gun,  and  the  height  of  the  muzzle 
above  the  plane  of  its  axis  if  the  muzzle  is  25  feet  long  from 
the  axis. 

Note  that  the  relative  positions  of  the  screens  are  usually 
determined  by  two  observers  in  towers  whose  distance  apart 
is  fixed ;  these  observers  record  positions  of  muzzle  and  each 
of  the  screens. 

23.  If  a  stick  of  length  12  units  casts  shadows  of  lengths  4, 
6,  8,  10,  12,  15,  18,  30,  and  40  units  respectively,  determine  the 
angle  of  inclination  of  the  sun.  For  angles  of  inclination  of 
10°  to  20°  by  degrees,  determine  the  corresponding  shadow 
length  to  tenths  of  one  unit.  This  type  of  table  was  the  first 
appearance  of  the  cotangent  function  as  direct  shadow ;  it 
appeared  as  early  as  900  A.D.  in  the  works  of  the  great  Arabic 

astronomer  Al-Battani. 

24.  The  pitch  of  a  roof  is  given 
by  the  vertical  height  li,  from 
the  point  C  to  M,  on  the  diagram, 
divided  by  the  span,  2  s ;  thus  ^ 
pitch  is  a  45°  slope.  Find  the 
slope  angle  of  a  roof  of  %  pitch, 

of  \  pitch.     If  2  s  is  given  as  48  feet,  find  the  length  of  the 

rafters  in  each  of  the  roofs  mentioned. 

25.  In  a  roof  of  span  62  feet  find  to  the  tenth  of  an  inch 
the  lengths  of  the  rafters  if  the  roof  is  inclined  at  30°,  40°,  42°, 
45°,  53°,  and  60°.     In  each  case  determine  the  effect  upon  the 
length  of  the  rafter  of  an  error  of  one  degree. 

26.  Find  the  pitch  and  the  angle  of  inclination  of  the  roof 
represented  in  the  diagram  above. 


Pitch  equals  —  when  span  is  2  s 
2s 


RIGHT  TRIANGLES 


217 


8.  Railroad  curves.  —  In  so  far  as  possible  the  track  of  a  rail- 
road is  laid  out  in  straight  lines.  Wherever  the  direction  of 
the  track  is  changed  a  curved  line  of  track  is- introduced  lead- 


Simple  curve  at  a  turnout  on  a  railroad  track 

ing  from  the  one  straight  track  to  a  second;  these  straight 
portions  of  track  must  be  tangent  to  the  curve  which  joins 
them,  and  so  they  are  commonly  designated  simply  as  tangents. 
Let  AV  and  VB  in  the  figure  represent  two  such  tangents, 
meeting  at  a  point  V,  called  the  vertex  ;  the  exterior  angle 
XVB  is  called  the  deflection 
angle,  and  is  usually  designated 
by  /.  A  single  circular  arc, 
radius  It,  which  joins  two  tan- 
gents is  called  a  simple  curve  and 
is  designated  in  American  rail-  Chord  100  feet;  arc  approxi. 
road  practice  by  the  number  of  de-  mately  100  feet 

grees  D  at  the  center  of  the  circle 

subtended  by  a  chord  whose  length  is  100  feet,  the  length  of 
one  chain  used  in  surveying.     On  a  simple  curve  the  lengths 


218  UNIFIED   MATHEMATICS 

of  two  consecutive  tangents,  from  intersection  point  to  the 
circle,  i.e.  AV  and  VB,  are  equal  ;  this  length  is  called  T. 
The  angle  D  is  commonly  given  only  in  degrees  and  half- 
degrees. 

Relation  between  D  and  R.  Let  PB  on  the  figure  represent 
a  chord  of  length  100  feet  ;  drop  the  perpendicular  from  A,  the 
center  of  the  circular  arc  of  radius  R,  bisecting  PB.  Evi- 

dently sin   —  =  —  ,  whence  R  =  —  —  .      Now  for  any  angle 
2      R  •    D 

sin  — 
2 

up  to  4°  the  sine  differs  numerically  from  the  angle  expressed 
in  radians  by  less  than  .1  of  1%  of  itself;  hence  you  may 

replace  sin  —  by  —  •  —  ^-  ,  the  value  of  —  in  radians,  with  an 
2         2     180  2 

error  of  less  than  .1°  of  1%  when  D  is  any  angle  up  to  8°. 
Note  that  the  error  is  less  than  o  feet  in  5000  ;  the  circular 
measure  of  the  angle  is  larger  than  the  sine  so  that  the  error 
will  be  a  deficiency. 

18000 

i\  ^  -— 

TTD 

gives  the  radius. 

Relation  between  I,  R,  and  T.     On  our  figure  in  the  right 

triangle  (MFthe  angle  AVO  is  90°  -  |,  and  the  angle  AOB  is 
evidently  equal  to  the  deflection  angle  /. 


-  , 

2J     AV     T' 

whence  R  =  T  cot  £, 

and  T  =  R  tan  f 

Evidently  the  radius  R  can  be  expressed  in  terms  of  the 
"degree"  D  of  the  curve,  giving  new  formulas  involving 
D,  T,  and  7. 

Elevation  of  outer  rail.  In  turning  a  curve  a  railroad  train 
tends  to  leave  the  track,  due  to  the  tendency  of  any  moving 
body  to  continue  its  motion  in  a  straight  line.  To  keep  the 
train  on  the  track  the  flanges  alone  are  not  sufficient,  but  the 


RIGHT  TRIANGLES  219 

outer  edge  must  be  elevated.    The  formula  for  ordinary  speeds, 

giving  number  of  inches  of  elevation,  is  e  =  -3- — ,  wherein  g  is 

32  R 

the  gage  of  the  track  in  feet,  v  the  velocity  of  the  train  in  feet 
per  second,  and  R  the  radius  of  curvature  in  feet. 


PROBLEMS 

1.  What  radii  have  railroad  curves  of  8°,  7°,  6°,  5°,  4°,  3°, 
2°,  and  1°,  respectively  ? 

2.  If  a  railroad  curve  is  built  with  the  radius  of  2640  feet, 
compute  D  in  degrees. 

3.  On  a  circular  track  of  100  miles'  circumference  what 
would  be  the  number  of  degrees  ? 

4.  On  English  and  continental  railroads  the  curvature  is 
usually  given  by  the  length  of  the  radius ;  find  the  number  of 
degrees,  American  D,  corresponding   to  radii  of   8000,  5000, 
4000,  3000,  2000,  1000,  800,  600,  and  400  feet,  respectively. 
Do  not  compute  beyond  minutes.     Find  D  for  radii  of  300 
meters,  1000  meters. 

5.  Compute  e,  elevation  of  outer  rail,   for  g  =  4   feet   8.5 
inches,  standard  gage   on  American    railroads,  when  v  =  60 
miles  per  hour,  and  R  =  800,  1000,  2000,  4000,  and  5000  re- 
spectively.    Compute  for  a  one-degree  and  for  a  two-degree 
curve. 

6.  Given  that  two  portions  of   straight   track   diverge   at 
22°  14',  and  that  the  tangent  distance  is  to  be  300  feet,  com- 
pute R ;  find  R  for  T,  the  tangent  distance,  equal  to  200,  250, 
and  350.     Find  the  corresponding  values  of  D.     How  could 
you  determine  the  length  of  T,  approximately  300,  so  that 
D  will  come  out  in  degrees  and  half-degrees  ? 

7.  Compute  R  when  T  =  400,  500,  and  600,  respectively, 
the  deflection  angle  being  60° ;  similarly  when  I  =  30°. 

8.  Compute  e  for  g  =  4  feet  8.5  inches  (4.71  feet),  standard 
gage,  v  =  60  miles  per  hour,  and  curves  of  1°,  2°,  5°,  6°,  and  8°> 
respectively. 


CHAPTER   XIV 
THE    CIRCLE 


1.    Formulas. — 

a-2 +  2 
(x-hy+(y-k 


==  r\      \x  —  h  =  r  cos  6,         Parametric  equa- 
=  r2.      |  y  —  Jc  =  r  sin  $.       tfons  of  the  circle. 


+  2/i2  =  r2  ;  X22  +  y22  =  r2  ;  x32  +  y32  =  r2  ;  x42  +  2/42  =  »"2  ;  x2  +  y2  =  r2. 


For   any  point  P  (x,  y)  on  a  circle  of   radius   10,  center  the 
origin,  we  have  the  relation, 

a;2  +  2/2  =  100, 
220 


THE  CIRCLE  221 

which  is  the  equation  then  of  a  circle  of  radius  10  and  center 
at  the  origin  (0,  0).  This  equation  is  obtained  directly  from 
the  distance  formula ;  x2  +  y2  =  100  expresses  the  fact  that  the 
distance  of  the  point  (x,  y)  from  the  point  (0,  0)  is  10 ;  any 
point  (x,  y)  which  satisfies  this  equation  is  at  a  distance  10 
from  (0,  0)  and  any  point  at  a  distance  of  10  units  from  O 
(0,  0)  satisfies  this  equation.  The  locus  of  this  equation,  then, 
is  the  circle  of  radius  10  and  center  (0,  0). 

The  formula  may  readily  be  verified  on  the  figure  ;  take 
P  any  point  on  the  circle,  drop  PM  a  perpendicular  to  the 
a?-axis.  Then  OM2  +  MP2  =  OP2,  in  any  one  of  the  four  tri- 
angles, representing  any  possible  position  of  P.  Herein  OM 
and  MP  must  be  regarded  initially  as  positive  quantities,  since 
the  formulas  of  plane  geometry  were  applied  only  to  positive 
lengths.  However  OM,  as  a  positive  length  =  #,  where  the 
negative  sign  is  taken  for  points  in  II  and  III  and  MP=y, 
where  the  negative  sign  is  taken  in  III  and  IV,  whence  sub- 
stituting in  OM*  +  MP2  =  OP2  you  have  xz  +  y*  =  102.  For  a 
circle  of  radius  r  the  equation  x2  +  y2  =  r1  is  satisfied  by  any 
point  P(x,  y)  which  is  upon  the  circle,  for  OP  will  equal  r,  and 
every  point  which  satisfies  the  equation  evidently  lies  on  the 
circle.  Hence,  by  definition,  the  locus  of  x2  +  y2  =  r2  is  the 
circle  of  center  (0,  0)  and  of  radius  r,  for  every  point  on 
the  circle  satisfies  this  equation  and  every  point  which  satis- 
fies the  equation  lies  upon  the  circle.  These  two  conditions 
must  be  fulfilled  in  order  that  any  given  curve  may  be  desig- 
nated as  the  locus  of  a  given  equation.  In  other  words,  the 
given  curve  must  include  all  points  whose  coordinates  satisfy 
the  equation  and  must  exclude  all  whose  coordinates  do  not 
satisfy  the  given  relation. 

Similarly  the  two  equations : 

x  =  10  cos  6, 
y  =  10  sin  0, 

give  for  every  value  of  0,  called  a  parameter,  the  coordinates 
of  a  point  which  lies  upon  the  circle.  The  locus  of  this  pair 


222 


UNIFIED  MATHEMATICS 


of  equations  is  the  circle  of  radius  10.  Thus  the  ten  values  of 
0=0°,  30°,  45°,  60°,  90°,  120°,_150°,  180°,  210°,  and  330°,  give' 
the  ten  points,  (10,  0),  (5\/3,  5),  (oV2,  5\/2),  (5,  5V3), 
(0,  10),  (-  5,  5V3),  (-  5V3,  5),  (-  10,  0),  (-  5V3,  -  5),  and 
(+  5V3,  —  5),  which  lie  upon  the  circle.  Intermediate  values 


Circle  of  radius  10  ;  units  are  eighths  of  an  inch 
f  X  =  10  cos  0. 


=  100,  or 


y  =  10  sin  6. 


can  readily  be  obtained  using  the  tables  of  sines  and  cosines. 

The  two  equations  together  constitute  the  equations  of  the 
circle  in  parametric  form,  a  type  of  equation  of  particular  im- 
portance in  applied  mathematics. 

If  desired,  we  may  eliminate  0  as  follows :  squaring  and 
adding  gives 

a-2  +  f  =  100  (cos2  8  +  sin2  6), 
or  x2  +  y>-  =  100  (since  sin2  0  +  cos2  0  =  1), 


THE  CIRCLE  223 

a  relation  independent  of  0.  But  for  many  purposes  it  is  more 
convenient  to  keep  the  equations  in  parametric  form. 

For  the  distance  from  any  point  C  (Ji,  A:)  to  a  point  P  (x,  ?/)  we 
have  found  the  formula  d  =  V(x  —  ft)2  +  (y  —  k)- ;  all  points 
(x,  y}  which  satisfy  this  equation  for  a  given  value  of  d,  and 
for  (h,  k)  a  fixed  point,  lie  upon  a  circle  of  which  (h,  k)  is  the 
center  and  d  is  the  radius  ;  no  point  not  on  the  circle  satisfies 
this  equation. 

(x  —  ft2)  +  (y  —  k}2  =  r2  is  then  the  equation  of  a  circle  of 
center  (h,  k)  and  radius  r.  Any  equation  which  can  be  put 
into  this  form  represents  a  circle,  for  it  expresses  the  fact 
that  the  distance  from  any  point  (x,  y)  whose  coordinates 
satisfy  the  given  equation,  to  the  fixed  point  (h,  k)  is  constant 
and  equal  to  r. 

In  parametric  form,  the  two  equations  representing  the 
circle  with  center  (ft,  k)  and  radius  r  are  written : 

x  —  ft  =  r  cos  0. 
y  —  k  =  r  sin  0. 

If  0  is  given  values  the  corresponding  values  of  x  and  y 
determine  points  upon  the  circle  (x  —  ft)2  +•  (y  —  fc)2  =  r2. 

Illustrative  problem.  —  Find  the  equation  of  the  circle 
of  radius  5 ;  center  (3,  —  7). 

By  the  distance  formula,  taking  (x,  y)  as  any  point  on  the  circle, 

(z- 3)2+(y  +  7)2  =  25, 

or  z2  -  6  x  +  y2  +  14  y  -  33  =  0. 

In  parametric  form  the  equations  of  this  circle  are, 
x  —  3  =  5  cos  0. 
y  +  7  =  5  sin  0. 

2.   Reduction  to  standard  form.  —  Any  equation  of  the  type, 

x2  +  /  +  2  Ox  +  2  Fy  +  C  =  0, 
or  Ax2  +  Af  +  2  Ox  +  2  Fy  +  C  =  0, 

represents  a  circle.  The  center  and  radius  are  determined  by 
completing  the  square,  as  in  the  illustrative  problem  below. 
If  the  expression  for  the  radius  is  zero,  the  circle  reduces  to  a 
point ;  if  it  is  negative  the  circle  is  imaginary. 


224  UNIFIED  MATHEMATICS 

Illustrative  problem.  —  Find  the  center  and  radius  of  the  circle, 


This  equation  represents  a  circle  since  it  can  be  put  into  the  form  of  a 
circle,  as  indicated  herewith  : 


2(x2  +  3  X  +  f  )  +  2(7/2  -  J  y  +  f  |)  =  15  +  £  +  -V. 
2(x  +  |)2  +  2(j/  -  |)2  =  *jp. 

(x  +  I)2  +  (y  ~  I)2  =  -¥65  (or  12.81). 

This  equation  states  that  the  point  (x,  y)  is  at  the  distance  --  from 

4 
the  point   (  -  f  ,  -J)  ;  this  equation  represents  a  circle  with  the  center 


(-§,!)  and  radius  -     -  or  -    -  or  3.58. 
4  4 

PROBLEMS 

Find  the  equations  of  the  following  circles : 

1.  Center  (3,  —  4),  radius  5.     3.    Center  (—  4,  0),  radius  4. 

2.  Center  (0,  0),  radius  10.       4.    Center  (-  6,  6),  radius  6. 

5.  Center  (—6,  —  8),  radius  10. 

6.  Draw  the  circle  of  radius  10,  center  (0,  0)  and  estimate 
carefully  its  area  on  the  coordinate  paper. 

Find  the  centers  and  radii  of  the  following  circles ;  time 
yourself ;  the  eight  problems  should  be  completed  numerically 
within  12  minutes. 

9.   x2  +  y2  -  39  =  0. 

12.  2  z2  +  2  y2  -  5  x  +  7  y  —  15  ==  0. 

13.  3  a;2  +  3  ?/2  -  15  x  +  IT  y  +  9  =  0. 

14.  x2  +  6  x  +  y2  -10  =  0. 

15.  Draw  the  graphs  of  the  preceding  8  circles,  using  only 
one  or  two  sheets  of  graph  paper;  time  yourself,  keeping  a 
record  of  the  time. 


THE  CIRCLE 


225 


16.  Given  x  =  5  cos  0, 

y  =  5  sin  6, 

locate  16  points  on  the  curve,  using  the  values  sin  0°  =  0, 
sin  30°  =  .5,  sin  45°  =  .707,  sin  60°  =  .866  for  these  and  related 
angles. 

17.  Given  x  =  5  +  5  cos  0, 

y  =  —  3  +  5  sin  0, 
locate  16  points  on  this  circle. 

18.  When  8  =  37°,  43°,  62°,  80°,  and  85°  find  x  and  y  in  the 
preceding  problems. 

19.  Through  what  point  on  the  circle  x2  +  yz=  25  does  the 
radius  which  makes  an  angle  arc  tan  2  with  OX,  pass  ? 


Photo  by  H.  J.  Karplnskl 


The  Rial  to  in  Venice 
A  famous  circular  arch,  95  feet  wide  by  25  feet  high. 

3.    To  find  the  intersection  of  a  line  with  a  circle.     Tangents.  — 

The  intersections  of  the  circle,  xz  +  yz  =  100,  with  any  line 
as  y  =  x  +  5,  are  represented  by  the  solutions  of  the  two  equa- 
tions regarded  as  simultaneous.  Six  problems  are  given  here. 

1.   a2 +  ?/2  =100,  2.   a2 +  y2  =  100, 

y  =  x.  y  =  x  +  5. 


226 


3. 


UNIFIED  MATHEMATICS 
00,  5. 


y  =  x.+  10. 

4.   x2  +  yz  =  100, 

y  =  a;  +  16. 


6.    a2  +  2/2  _  100, 
y  =  a  -f  fr. 


Solving,  by  substitution  in  each  of  the  six  cases  indicated 
above : 


Olr 


£ 


Graphical  solution,  determining  intersections  of  the  circle,  x2  +  y2 
with  various  lines  of  slope  1 

1.  gives  2  x2  =  100,  x2  =  50,  x  =  ±  \/50  =  ±  7.07,  y  =  ±  7.07  ; 

2.  gives2x2  +  10x  +  25  =  100,  x2  +  5x-37.5  =  0; 

x  =  -  2.5  ±  V6.25  +  37.5  =  -  2.5  ±  6.61 

=  +  4.11  or  -9.11, 
i/  =  9.11  or  -4.11; 

3.  gives  x2  +  10 x  =  0,  x  =  0  or  —  10  (by  factoring,  simplest), 

y  =  10  or  0  ; 


THE  CIRCLE  227 


4.    gives  2  x2  +  32  x+  156  =  0,  x2  +  16x  +  78  =  0, 

-  78 


=  _  8  +  V-14 

=  imaginary  values,  not  any  scalar  values  of  x; 

5.  gives  2  x2  -  16x  -  36  =  0,  x2  -  8  x  -  18  =  0,  x  =  4  ±  V34 

=  4  ±  5.83 
=  9.83,  or  -  1.83, 
y  =  1.83  or  -  9.83  ; 

6.  gives  2  x2  +  2  fc  •  x  +  (k2  —  100)  =  0.  _ 

-  2(fc2  -  100) 


Very  evidently  the  solutions  of  1  to  5  are  all  included  under 
the  solution  6,  as  special  cases. 

Geometrically  the  lines  of  slope  1  are  divided  by  the  circle 
into  three  classes,  viz.  (a)  those  which  cut  the  circle  in  two 
distinct  points  ;  (&)  those  which  do  not  cut  the  circle  ;  and  (c) 
those  which  are  tangent  to  the  circle,  or  cut  the  circle  in  two 
coincident  points. 

Evidently  lines  in  1,  2,  3,  and  5  belong  to  the  first  class  ;  the 
line  in  4  to  the  second  class.  To  determine  the  tangents  one 
must  find  the  value  of  k  for  which  200  —  k2  =  0,  as  only  when 
200  —  k2  =  0  are  two  points  whose  abscissas  are  given  by 

-  k  +  *  V200  -  fc2  and  -  *  -  -  V200  -  fc2  coincident.     In  this 

a        a  a        u 

case,  k  =  ±  14.14,  the  lines  y  =  x  ±  14.14  are  tangent  to  the 
circle  a;2  +  y2  =  100.  The  abscissa  of  the  point  of  tangency  is 

A; 
T  7.07,  since  it  equals  --- 

^ 

KULE.  —  To  find  the  tangent  with  given  slope  to  a  given  circle 
write  the  equation  of  the  family  of  lines  of  the  given  slope, 
y  =  mx  +  k,  and  solve  for  the  points  of  intersection  with  the  circle; 
get  the  condition  that  the  two  points  of  intersection  should  be  coin- 
cident. This  gives  the  value  of  k  for  which  the  line  y  =  mx  +  A: 
is  tangent  to  the  given  circle. 

NOTE.  —  The  method  will  apply  to  any  curve  of  the  second  degree. 


228 


UNIFIED  MATHEMATICS 


4.  Circles  satisfying  given  conditions.  —  To  find  the  equation 
of  a  circle  which  satisfies  given  conditions  it  is  necessary  to  use 
the  analytic  formulas  which  we  have  derived  combined  with 
the  geometric  properties  of  a  circle.  In  general  call  (h,  k)  or 
(x,  y)  the  center  of  the  circle  and  r  the  radius ;  sketch  the  lines 
and  points  which  are  given  and  indicate  roughly  the  probable 
position  of  the  desired  circle ;  solve  the  problem  geometrically 
if  possible,  or  indicate  the  solution,  and  express  the  geometrical 
facts  in  algebraical  language  by  using  the  preceding  formulas. 


Circle  through  three  points 
Determination  of  center  by  perpendicular  bi- 
sectors of  chords. 


5.  Illustrative 
problems.  —  Find 
the  equation  of  the 
circle  through  A 
(1,  2),  B  (0,  8),  and 
C  (7,  -  1),  (1)  using 
the  distance  formula, 
(2)  using  the  per- 
pendicular bisector 
of  the  line  joining 
two  points,  (3)  using 
the  general  equation 

(X  — fc)2+(y  — &)8=r* 

which  may  represent 
any  circle,  and  (4) 
using  the  general 
equation 

Ax2  +  Af  +  2Gx 


(1)  Call  the  center  P(h,  *),  then  PA=PB,  PB  -  PC,  and  PA=PC. 

The  distance  from  A  to  P  equals  the  distance  from  B  to  P,  whence  by 
the  distance  formula, 


-  1)2  +  (*  -  2)2  =  V(h  -  0)2  +  (k  -  8)2  ; 


THE  CIRCLE  229 


Similarly,   v 

II.  ^/(h-  l)2+(*-2)2  =  v'(h  -  7) 

expresses  analytically  the  fact  that  PA  =  PC  ;  and 


III.  V(&-0)2  +  (Jk-8)2=  \/(A-7)2  +  (Jfc  +  1)2 

that  PS  =  PC. 

Since  equation  III  is  derivable  from  I  and  II,  it  adds  nothing  new  ; 
any  two  of  these  equations  are  sufficient  to  determine  (7t,  &)  the  center. 
Squaring  in  each  and  combining  terms  we  obtain  from  I, 

12  k  -  2  h  -  59  =  0, 

a  straight  line  which  is  the  locus  of  all  points  equidistant  from  A  and  B  ; 
and  from  II,  4  A  -  2  fc  -  15  =  0. 

Solving,  we  obtain  the  one  point  which  is  equidistant  from  A,  B,  and  C, 
h  =  6.77. 

k  =  6.05.  _         _        _ 
r  =  V(6.77  -  I)2  +  (6.05  -  2)2  =  V(33.29  +  16.40)  =  V49709 

=  7.05. 
The  circle  is  (x  -  6.77)2  +  (y  -  6.05)2  =  (7.05)2. 

(2)  The  center  of  the  circle  is  the  intersection  of  the  perpen- 
dicular bisectors  of  the  sides  ;  finding  the  slopes  of  the  sides, 
the  mid-points,  the  slope  of  the  perpendicular  to  each  side,  the 
equations  of  the  perpendicular  bisectors  of  AB  and  AC  are 
found  (point-slope)  to  be 

12  y  -  2  x  -  59  =  0. 
4x-2y  -15=0. 

Examination  shows  that  these  are  precisely  in  x  and  y  the  equations 
obtained  in  our  first  solution  in  h  and  k  and  from  this  point  the  solution 
proceeds  as  in  (1).  The  student  should  explain  the  reason  for  this. 

(3)  I.  (x  -  A)2  +  (y  -  A:)2  =  i* 
is  the  equation  of  any  circle,  center  (7i,  ft),  radius  r. 

Substituting  in  this  equation  (1,  2),  (0,  8),  and  (7,  —  1),  gives, 

II.  (1  -  A)2  +  (2  -  ky  =  r2. 

III.  (0  -  A)2  +  (8  -  Jfc)«  =  r2. 

IV.  (7  -  hy  +  (-  1  -  fc)2  =  r2. 

V.  Ill  -  II  -  2  h  +  12  A;  -  59  =  0. 

VI.  II  -  IV  -  48  +  12  h  +  3  -  6  k  =  0 

or  4  h  -  2  k  -  15  =  0. 


230  UNIFIED  MATHEMATICS 

V  and  VI  are  seen  to  be  in  //  and  k  pivdsrly  tin-  equations  solved  in 

method  (1).  and  in  method  (-2)  for  x  and  //  ;us  variables. 

(4)  I.  Ac*  +  Ayn-  +  2  Gx  +  2  Fy  +  C  =  0. 

Substitute  in  this  equation  (1,  2),  (0,  8),  and  (7,  —  1)  and  solve  for 
the  values  of  G,  F,  and  C  in  terms  of  A. 

II.  A  +  4A  +  2G  +  4F+C  =  0. 

III.  64  A  +  16  F  +  C  =  0. 

IV.  49^1  +  A  +  UG-2F+  C  =  0. 

V.  II  -  III  -  59  A  +  2  G  -  12  F  =  0. 
-VI.    II  -IV  -45A+12G  +6F  =  0 


C1  F 

These  are  the  same  equations  in  --   and  --  ,  regarded  as  the  un- 

A  A 

knowns,  as  appeared  above  in  h  and  k. 

6.  Tangency  conditions.  —  If  a  circle  is  to  be  tangent  to  a 
given  line  the  distance  formula  (normal  form)  from  a  point  to 
a  line  may  be  used  ;  if  a  circle  to  be  found  is  to  be  tangent  to 
a  given  circle,  then  the  radius  sought,  plus  or  minus  the  given 
radius,  must  be  equal  numerically  to  the  distance  from  center 
to  center,  according  as  the  circles  are  tangent  externally  or 
internally. 

7.  Circle  through  the  intersection  of  two  circles.  — 

I.  x2  -+-  yz  +  10  #  =  0,  a  circle  of  radius  5,  center  (—5,  0). 

II.  x*  +  yz  —  40  =  0,  a  circle  of  radius  7,  center  (0,  0). 

III.  (x2  +  y-  +  10  JT)  +  A-(x2  +  2/!  -  49)  =  0. 

The  third  equation  is  satisfied  by  the  points  of  intersection 
of  curves  I  and  II,  for  all  values  of  A:  (see  page  83).  For  all 
constant  values  of  k,  III  may  be  written 


(1  +  k)x*  +  (1  +  k)y*  +  lOx  -  49  k  =  0, 

and  the  form  shows  that  this  represents  a  circle.  To  deter- 
mine the  circle  through  the  intersections  of  I  and  II,  and  any 
other  given  point  substitute  the  coordinates  in  III,  and  solve 


THE  CIRCLE 


231 


for  A- ;  since  a  circle  is  determined  by  the  three  points,  it  is 
easily  seen  that  every  circle  through  the  two  points  of  inter- 
section of  the  given  circle  is  included  in  the  family  of  circles, 
(1  +  fr)  a*  +  (1  +  A-)  y2  -f  10  x  -  49  k  =  0.  The  method  of  deter- 
mining A:  to  have  the  circle  pass  through  some  other  given 


Common  chord  of  two  circles  or  radical  axis 

point  is  precisely  the  same  as  in  the  similar   problem  with 
straight  lines  (page  83). 

For  A'  =  —  1,  this  equation  reduces  to  the  linear  equation 
representing  the  common  chord  of  the  family  of  circles ; 
whether  two  given  circles  intersect  or  not,  this  line,  whose 
equation  is  obtained  by  eliminating  x2  +  y2  between  the  two 
given  equations,  is  called  the  radical  axis  of  the  two  circles. 

8.   Geometrical  property  of  the  radical  axis.  — 

(x  —  h)°  +  (y  —  A-)2  is  the  square  of  the  distance  from  (x,  y) 
to  the  center  of  any  circle ;  (a;  —  A)2  •+-  (y  —  k)-  —  r*  is  the 
square  of  the  length  of  the  tangent  to  the  circle  from  any 
point  outside  the  circle  of  center  (h,  k),  radius  r. 

x*  _j_  y*  .j.  2  Gx  +  2  Fy  +  C  is  the  square  of  the  length  of  the 


232 


UNIFIED  MATHEMATICS 


tangent  from  any  point  (x,  y)  to  the  circle  whose  equation  ia 
x2  +  y2  +  2  Gx  +  2  /fy  +  C=  0,  since  the  left-hand  member  is 
identical  with  the  left-hand  member  when  written  in  this 
form: 

(x  +  £)2  +  (y  +  F)*-(G*  +  F*-  C')=  0. 

Note  that  if  any  secant  PAB  is  drawn  through  P(x,  y)  then 
PA  •  PB  =  -PT2;  hence  the  expression 

C 


gives  the  product  of  the  two  distances  along  any  straight  line 
from  the  point  P(x,  y)  on  the  line  to  the  circle.     There  is  a 


Distances  from  a  point  to  a  circle 

On  any  secant  through  P,  PAB,  PA  x  P.B  is  constant. 
PA  x  PB  =  P2V2  =  (x  -  hy  +(y-  fc)2  -  i*. 


correspondence  to  the  normal  form  of  a  straight  line,  since  the 
left-hand  member  there  also  represents  a  distance. 


is  an  equation  which  is  satisfied  by  any  point  from  which  tan- 
gents drawn  to  the  two  circles 


a*  +  f-  +  2  G&  +  2  F$  +  C,  =  0, 
x*  +  y*  +  2  G&  +  2  Fzy  +  C2  =  0, 


THE  CIRCLE 


233 


are  equal  in  length.  Hence,  the  radical  axis  is  the  locus  of 
points  from  which  the  tangents  drawn  to  the  two  circles  are 
equal  in  length. 

9.    Radical  center  of  three  circles.  —  Given  three  circles,  each 
of  the  three  pairs  of  circles  which  may  be  formed  from  the 


Radical  axes  and  radical  centers 
Radical  center  of  the  three  circles,  1,2,  and  3. 
Radical  center  of  the  tjiree  circles,  1,  2,  and  3'. 

three  has  a  radical  axis  ;  the  three  radical  axes  pass  through  a 
common  point,  as  may  be  easily  shown  by  Sec.  4,  Chapter  V. 
a.   X2  +    2  + 


b. 
c. 

d.  (a-c) 

e.  (b-c) 


radical  axis  of  a  and  b. 

2-F3)y+C»-C3= 

radical  axis  of  b  and  c. 


234  rXIFIKI)    MATIIKMATICS 


/.    (d  +  «)or(a-c)     2(G1- 

9  ice  d  +  e  =  0  gives  a  straight  line  through  the  intersec- 
tion of  d  and  e,  and  since  d  +  e  =  0  gives  the  radical  axis  of  a 
and  c,  the  latter  line  passes  through  the  intersection  of  the 
two  former  radical  axes. 

10.     Limiting  forms  of  the  circle  equation.  — 

(x  —  h)-  +  (y  —  k)-  =  r-  represents  a  real  circle  when  r2  is 
positive. 

(x  —  K)2  -\-(y  —  A-)2  =  0  represents  a  point  circle;  the  only 
real  point  which  satisfies  this  equation  is  the  point  (h,  k). 

(x  —  }()-  +  (y  —  k)-  =  —  r2,  r  a  real  quantity,  represents  an 
imaginary  circle  :  no  real  point  satisfies  this  equation,  since 
every  real  value  of  x  and  y  makes  (x  —  h)2  positive  and  (y—  k)2 
positive. 

PROBLEMS    ON    THE    CIRCLE 

1.  Find  the  center  and  give  radius  to  1  decimal  place  of 
each  of  the  following  circles  ;  plot  ;  find  the  three  radical  axes 
and  the  radical  center. 

a.  a?  +  .v2  +  6x  —  8y  —  16  =  0. 

b. 


HINT.  3(z2  —  j  x  )  +  3(j/2  +  5  y  )  =  7  ;  complete  squares  inside 
parentheses  and  note  that  3  times  the  quantity  added  within  each  of  the 
parentheses  must  be  added  on  the  right. 

c.  2?  +  y2-  6  £-8  =  0. 

2.  Plot    the    following    two   circles   and   determine    their 
common  chord  :  what  is  its  length  '.' 

a.   *»  +  y*  -  10  x-  100  =  0. 
6.    ^  +  ^  +  10^-100  =  0. 

3.  Write  the  equation  of  the  family  of  circles 

a.  With  center  on  ;c-axis,  passing  through  the  origin. 
//.  With  center  on  y-axis,  passing  through  the  origin. 
c.  Passing  through  the  origin. 


THE  CIRCLE  235 

d.  With  center  on  3  x  —  -i  y  —  5=0,  radius  5. 

NOTE.       3  h  —  4  k  -  5  =  0. 

4.  What  limitation  is  imposed  upon  the  coefficients  A,  G, 
JP,  and  07 in      ^  +  ^  +  2  Gfc  + 21fy+ 07=0,  • 

a.  if  the  circle  passes  through  (0,  0)  ?    (1,  1)  ? 

b.  if  the  circle  has  its  center  on  the  axis  of  x  ?  y-axis  ? 

c.  if  the  circle  is  tangent  to  the  aj-axis  ?  y-axis  ?  tangent  to 
z-3  =  0? 

d.  if  the  circle  is  tangent  too;  —  y  —  5  =  0? 

5.  Find  the  equations  of  the  circles  through  the  following 
three  points : 

a.  (0,  0),  (6,  0),  (0,  8). 

b.  (1,5),  (-3,1),  (7, -3). 

c.  (0,  0),  (8,  2),  (15,  —  3) ;  use  two  different  methods. 

6.  Find  circle  tangent  to  3  x  +  4:  y  —  25  =  0,  and  passing 
through  (2,  3)  and  (5,  1).     Note  the  two  solutions. 

7.  Find  the  radical  axis  of   each  of  the   three   pairs  of 
circles  a»  +  .y*- 6s -8y- 10  =  0, 

a-2  +  y2  -  20  x  +  50  =  0, 
2y?  +  2 yz  4-  Sx  +  6y  -  25  =  0. 
Find  the  radical  center.     Plot. 

8.  Find  the  tangents  of  slope  2  to  the  first  circle  in  7 ;  find 
the  normal  and  the  point  of  tangency. 

9.  Find  the  circle  of  radius  5  tangent  to  the  line  whose 
equation  is  4  x  —  3  y  —  9  =  0  at  (3,  1). 

10.  Find  for  what  value  of  r  the  line  4#  —  3?/  —  9  =  0  is 
tangent  to  y?  +  y2  —  r2  =  0.     Two  methods.     Find  the  point  of 
tangency. 

11.  Find  to  one  decimal  place  the  points  of  intersection  of 
the  circle  y?  +  y2  —  20  x  +  50  =  0  with  the   line  y  =  2  x  -  12. 
Plot. 


236  UNIFIED  MATHEMATICS 

12.    Use  the  trigonometric  functions  to  find  points  of  inter- 
section of 


Note  that  tan  0  =  2,  where  0  is  the  slope-angle  of  the  line. 

13.  Use  trigonometric  functions  to  find  k,  when  y  =  2  x  +  k 
is  tangent  to  the  circle  x2  +  y2  =  100.     Draw  figure  ;  note  that 
tan  0  =  —  |  where  0  is  the  slope-angle  of  the  normal. 

[  x  =  3  4-  10  cos  e, 

14.  Plot  the  circle  <  e      .,„    .     ., 

1  1/  =  —  5  +  10  sin  Q. 

[  x  =  8  sin  0, 

15.  Plot  the  circle  { 

[  ?/  =  8  cos  0. 

Note  that  0  is  here  the  angle  made  with  the  j/-axis  by  any  radius. 

16.  Find  the  equation  of  the  complete  circle  of  the  circular 
arch  of  the  Rialto,  referred  to  the  horizontal  water  line  and 
the  axis  of  symmetry  of  the  arc  as  axes.     The  arch  is  95  feet 
wide  by  25  feet  high. 

17.  Find  the  equation  of  the  circle  of  which  the  arch  of  the 
Eocky  River  Bridge,  280  feet  by  80  feet,  is  an  arc,  referred  to 
a  tangent  at  the  highest  point  of  the  arc  as  avaxis  and  the 
perpendicular  at  the  point  of  tangency  as  y-axis.     Determine 
the  lengths  of  vertical  chords  between  the  arc  and  the  x-axis, 
spaced  at  intervals  of  forty  feet. 


CHAPTER   XV 


ADDITION   FORMULAS 

1.    Functions  of  the  sum  and  difference  of  two  angles.  —  The 

formulas  for  (a  +  6)2  and  (a  —  6)2  are  illustrations  of  addition 
formulas  frequently  of  fundamental  importance  in  mathe- 
matical work.  Thus  10*  •  10"  =  10*+"  is  an  addition  formula 
leading  to  the  whole  theory  of  logarithms,  which  revolutionized 
computation  processes.  The  question  arises  as  to  addition 
formulas  in  the  case  of  the  trigonometric  functions  after  the 
functions  have  been  defined.  Just  as  the  exponent  formula 
1Qz+v  — 10* .  10",  which  was  first  proved  for  positive  integers, 
is  extended  to  hold  for  all  values  of  x  and  y,  so  the  formulas 
which  are  established 
for  sin  (a  +  ft)  and 
cos  (a  +  ft)  when  a 
and  ft  are  acute 
angles  will  be  found 
to  hold  for  all  real 
values  of  a  and  . 


f 


.V 


2.  Geometrical  der- 
ivation of  sin  (a  +  P) 
and  cos  (a  +  P) ;  a 
and  (J  acute  and 
a  +  p<90°.  —  Given 
a  and  ft,  two  acute 
angles  whose  sum  is 
less  than  90°,  to  find 

sin  (a  +  ft)  and  cos  (a  +  /?)  in  terms  of  sin  a,  cos  a,  sin  ft,  and 
cos  ft. 

237 


OB  =  r 


238 


UNIFIED   MATHEMATICS 


On  the  figure  let  a  and  ft  be  two  positive  acute  angles  whose 
sum  is  less  than  90°,  taken,  for  convenience,  distinctly  differ- 
ent from  each  other.  Let  OP  make  the  angle  <*  with  OX,  and 
OB  make  the  angle  ft  with  OP,  and  thus  «  +  ft  with  OX. 
From  B  drop  perpendiculars  BA  to  OP  and  BN  to  OX  ;  from 
A  on  OP  drop  a  perpendicular  AM  to  OX;  from  A  draw  a 
parallel  to  OX  cutting  BN  at  R. 

On  the  figure,  noting  that  OB  is  taken  as  r,  we  have  the 
following  evident  relations  : 

AB  =  r  sin  ft  ;  .R.B  =  J.J5  cos  a 

=  r  cos  a  sin  /3  ; 

OA  =  rcos/3-,  AM=  OA  sin  a 
=  r  sin  a  cos   ? 


=  r  sin  a  sin  ft  ; 
—  OA  cos  a=r  cos  a  cos  ft. 


sm 


_  r  sin  a  cos  ft  +  r  cos  «  sin  /? 


sin  (a  -f-  ft}  =  sin  «  cos  ft  +  cos  a  sin  ft. 
Similarly,  cos  (.  +  »  .  ™T= 


whence 


_  r  cos  a  cos  ft  —  r  sin  a  sin  ft 

t 
r 

cos  (a-\-  ft)  =  cos  a  cos  ft  —  sin  a  sin  ft. 


Having  established  these  formulas  geometrically  for  a  -f  ft 
when  0  <  a  <  90°,  0  <  ft  <  90°,  and  «  +  0  <  90°,  it  now 
remains  to  establish  that  these  formulas  hold  for  all  angles 
a  and  ft,  including  negative  angles.  This  extension  is  made 
by  employing  the  theorems  of  Section  12,  Chapter  VII. 


ADDITION  FORMULAS  239 

3.    Generalization  for  any  two  acute  angles.  — 

sin  («  +  ft)  =  sin  a  cos  ft  4-  cos  a  sin  ft, 
cos  (a  +  ft)  =  cos  «  cos  y8  —  sin  a  sin  /!?. 

First  we  will  show  that  when  a  and  ft  are  cmy  two  acute 
angles  the  two  formulas  established  above  when  a  +  ft  <  90° 
continue  to  hold.  The  extension  to  any  acute  angles  requires 
that  we  prove  these  formulas  to  be  true  further  (a)  when 
«  +  ft  =  90°,  and  (b)  when  «  +  ft  <  90°. 

Proof,     (a)  If  «  +  /8  =  90°,  /?  =  90°  —  a,  whence 
sin  /?  =  sin  (90°  —  a)  =  cos  a  ;  cos  ft  =  cos  (90  —  a)  =  sin  a. 
The  two  formulas  then  give,  by  substitution, 

sin  («  +  /?)=  sin  90°  =  sin2  a  +  cos2  a  =  1, 

cos  (a  +  /?)  =  cos  90°  =  cos  «  sin  a  —  sin  a  cos  «  =  0. 

The  sine  of  90°  is  1,  and  the  cosine  of  90°  is  0 ;  hence  our 
formulas  continue  to  hold  even  when  a  -f-  ft  =  90°. 

(6)  a  +  ft  >  90°.  Take  the  complements  of  a  and  /J  to  be 
respectively  x  and  ?/,  whence  x  =  90°  —  «  and  y  =  90°  —  ft. 
Evidently  x  +  y  will  be  less  than  90°,  by  the  same  amount  that 
a  +  ft  exceeds  90°.  Further,  since  x  =  90°  —  a  and  y  =  90°  —  ft, 

sin  x  =  cos  a,  sin  y  =  cos  /8, 

cos  x  =  sin  «,  and  cos  y  =  sin  ft. 

Now,  sin  («  +  ft)  =  sin  (90°  -  x  +  90°  -  y) 

=  sin  (180°  —  x  +  y)  =  sin  (a;  +  2/)>  since 
sin  (180°  — 0)=  shift 

Since  x  +  y  <  90°,  sin  (a?  +  y )  =  sin  cc  cos  y  +  cos  a;  sin  y,  as 
established  above ;  making  the  substitutions  for  sin  x,  cos  y, 
cos  x,  and  sin  y,  we  have  sin  (a  +  /?)  =  cos  a  sin  /?  -f-  sin  a  cos  /J. 

Q.  E.  D. 

Similarly, 

cos  («  +  ft)  =  cos  (180°  —  x  +  y)  =  —  cos  (x  +  y),  since 
cos  (180°  —  0)  =  —  cos  6,  for  any  angle  6.  But  x  and  y  are 
acute  angles,  whose  sum  is  less  than  90° ; 


240  UNIFIED  MATHEMATICS 

therefore  cos  (x  +  y)  =  -f-  cos  x  cos  ?/  —  sin  x  sin  y 

=  4-  sin  «  sin  /?  —  cos  a  cos  /?. 

Now  cos  (a  +  ft)  =  —  cos  (x  +  y), 
or  cos  (a  +  ft)  =  cos  a  cos  /3  —  sin  «  sin  /?.  o.  K.  i>. 

4.  Extension  of  the  formulas  for  sin  (a  +  0)  and  cos  (a  +  P)  to 
all  angles  without  restriction.  —  To  show  that  these  formulas 
hold  for  all  angles  it  is  necessary  now  to  show  that  if  either  a 
or  ft  is  increased  by  90°  the  formulas  continue  to  hold  pro- 
vided that  they  hold  for  a  and  ft. 
Thus  given 

sin  (a  +  ft)  ='  sin  a  cos  ft  +  cos  a  sin  ft, 
cos  (a  +  ft)  =  cos  a  cos  ^8  —  sin  «  sin  ft, 

we  wish  to  show  that  sin  («  -f-  ?/)  and  cos  (a  +  y~)  are  given  by 
similar  formulas,  when  y  =  ft  +  90°. 

sin  (a  +  y)=  sin  («  +  ft  +  90°)  =  cos  («'+  /?), 
=  cos  a  cos  ft  —  sin  a  sin  ft, 

but  sin  ?/  =  cos  ft  and  cos  y  =  —  sin  /?,  whence  substituting, 

sin  (a  -f-  y)=  cos  a  sin  y  +  sin  a  cos  y.  Q.  E.  D. 

Similarly  for  cos  («  +  ?/),  we  find  cos  a  cos  y  —  sin  a  sin  y ; 
since  a  and  /?  enter  symmetrically  in  the  above  formulas  this 
•proof  establishes  that  a  also  could  be  increased  or,  by  an 
entirely  analogous  procedure,  decreased  by  90°,  with  the  same 
formulas  for  the  new  values. 

This  establishes  the  formulas  for  any  two  angles  a  and  ft 
whatever.  For  since  the  formulas  have  been  proved  above  to 
hold  for  any  two  acute  angles  «  and  ft,  the  formulas  hold  for 
any  obtuse  angle  and  any  acute  angle  since  y,  the  obtuse  angle, 
may  be  regarded  as  90°  +  ft.  This  establishes  the  formulas  for 
any  angle  in  I  and  any  angle  in  II ;  now  increase  a  by  90°, 
thus  establishing  the  formula  for  any  two  obtuse  angles. 
Continuing  in  this  way  «  can  be  any  angle  in  any  quadrant,  I 
to  IV,  and  ft  also  an  angle  in  any  quadrant  whatever,  and  the 
formulas  continue  to  be  true. 


ADDITION  FORMULAS 


241 


After  these  formulas  are  established  for  all  positive  angles 
up  to  360°,  another  method  of  procedure  to  establish  the 
formulas  for  all  positive  and  negative  angles  is  to  note  that 
any  integral  multiple  of  360°,  k  •  360°  with  k  a  positive  or 
negative  integer,  can  be  added  to  any  angle  without  changing 
the  functions  of  the  angle  involved  in  our  formulas.  Thus  if 
—  ft  is  any  negative  angle,  numerically  less  than  360°,  the 
functions  of  a  +  (—  ft)  are  the  same  as  the  functions  of 
a  +  (360°  —  ft)  which  is  the  sum  of  two  positive  angles  ;  but 
the  functions  of  360°  —  ft  are  the  same  as  those  of  —  ft  and 
after  application  of  the  formula  the  360°  can  be  dropped.  In 
other  words  in  these  formulas  any  integral  multiple  of  360° 
can  be  added  at  pleasure  and  also  dropped  at  pleasure,  and  in 
this  way  the  formulas  are  established  for  all  angles. 

Illustrative  problem.  —  Given  sin  a  —  .45,  cos  ft  =  .68,  find 
sin  (a  +  /?)  and  cos  (a  +  ft). 

sin  a  =  .45  ;  a  can  be  in  I  or  II  since  sin  (180°  —  «)  =  sin  a. 
cos  a  =  ±  Vl  -  .452  =  ±  V/7975  =  ±  .893. 
cos  ft  =  .68 ;  ft  can  be  in  I  or  IV  since  cos  (—  ft")  =  cos  ft. 
sin  ft  =  ±  VI  -  .682  =  V(.32)(1.68)  =  ±  .16  x  V21 
=.  ±  .16  x  4.58  =  ±  .733. 


sin  a.  -  .45  determines  either 
a,  or  a-.> 


cos  (3  =  .68  determines  either 
Pi  or  p2 


There  are  strictly  four  problems,  solved  as  follows  : 

a  in  I,  ft  in  I.  a  in  I,  ft  in  IV. 

sin  «  =  .45. 
cos  a  =  +  .893. 
cos  ft  —  +  .68. 


242  UNIFIED   MATHEMATICS 

sin  £=  +  .733.  sin  £  =  -.733. 

sin(a  +  £)  sin(«  +  £) 

=  sin  a  cos  £  +  cos  a  sin  £.  =  .45  x  .68  —  .893  x  .733. 

sin  (a  +  £)  sin  («  +  £) 

=  .45  x  .68  +  .893  x  .733  =  .306  -  .655  =  -  .349. 

=  .306  +  .655  =  .961.  cos  (a  +  £) 
cos  (a  +  £)  =  cos  a  cos  £  —  sin  a  sin  £. 

=  .893  x  .68  -  .45  x  .733  cos  (a  +  £) 

=  .607  -.  .330  =  .277.         =  .893  x  .68  +  .45  x  .733 

=  .607  +  .330  =  .937. 

The  two  columns  represent  two  solutions  which  have  the  three 
central  values,  sin  a,  cos  a,  and  cos  ft,  in  common. 

The  student  is  expected  to  complete  the  solution,  beginning 
as  follows : 

a  in  II,  £  in  I.  a  in  II,  £  in  IV. 

sin  a  =  .45. 

cos  ft  =  .68. 
cos  «  =  -  .893. 
sin  £  =  +  .733.  sin  £  =  -.733. 

In  general  work  only  one  case,  indicating  which  solution  is 
given. 

5.  Historical  note.  —  The  formulas  for  sin(a  +  £)  and 
cos(«+£)  are  closely  allied  to  Ptolemy's  theorem  (c.  150  A.D.) 
that  in  any  inscribed  quadrilateral  the  product  of  the  diagonals 
is  equal  to  the  sum  of  the  products  of  the  opposite  sides.  If  a, 
6,  c,  and  d  are  the  sides,  in  order  around  the  quadrilateral,  and 
e  and /the  diagonals,  e/=oc  +  6tZ;  in  the  Greek  trigonometry 
employing  chords  this  theorem  plays  the  same  role  that  the 
formulas  for  sin  (a  +  £)  and  cos  (a  +  £)  play  in  the  trigonometry 
employing  sines  and  cosines.  A  great  French  mathematician, 
Viete  (1540-1603),  the  first  to  use  generalized  coefficients  in 
algebraic  equations,  was  the  first  to  give  these  formulas,  as 
sin  (2  a  +  £)  and  cos  (2  a  +  £)  in  terms  of  a  +  £  and  a ;  the 


ADDITION  FORMULAS  243 

modern  form  appeared  in  1748  in  the  work  of  the  Swiss  mathe- 
matician Euler. 

PROBLEMS 

1.  Given  a  =  30°,  j8  =  45°,  find  sin(«  +  £)  and   cos(a-f /3). 
Check  by  tables. 

2.  Given  a  =  60°,  p  =  45°,  find  sin  105°,  and  cos  105°.    Check 
by  the  preceding  problem,  and  explain  the  check. 

3.  Given  sin«  =  f,  and  sin  p  =  ^}  find  sin  («  +  /?),  when  a 
and  p  are  both  acute ;  find  sin  (a  +  /3)  when  a  and  p  are  both 
obtuse ;  when  a  is  obtuse,  /?  acute. 

4.  Given  a  and  p  acute   angles,  sin  a  =  .351,  cos  p  =  .652, 
find  sin  («  -+-  /3)  by  the  formula  and  check  with  the  tables. 

5.  Given  sin  18°  =  .3090,  cos  18°  =  .9511,  find  sin  36°. 

6.  Given  sin  18°  =  .3090,  find  sin  78°. 

7.  Using  the  results  of  problem  1  for  sin  75°  and  cos  75° 
with  the  data  of  problem  5,  find  sin  93°  and  cos  93°  ;  thus  find 
sin  3°  and  cos  3°. 

8.  What  are  sin  (45°  +  a)  and  cos  (45°  +  «)  in  terms  of  a  ? 

9.  Find  sin  (60°  +  a)  in  terms  of  sin  a  and  cos  a. 

6.  The  formulas  for  sin  (a  —  p)  and  cos  (a  —  P).  —  If  (3  is  a 
negative  angle,  a  —  p  comes  directly  under  the  a  +  /3  formula 
as  «  H-  ( —  (3) ;  if  (3  is  any  positive  angle  greater  than  360°,  p 
can  be  reduced  to  less  than  360°  by  subtracting  360°  (or  some 
multiple  of  360°)  without  affecting  the  functions  of  «  —  ^8  or  of 
p ;  if  (3  is  positive  and  less  than  360°,  the  functions  of  «  —  ft 
will  be  the  same  as  the  functions  of  a +(360°  —  (3),  since  this 
simply  adds  one  complete  revolution  to  a  —  p.  Hence 
sin  («  —  p)  = 

sin  («  +  360°  -  P)  =  sin  a  cos  (360°  -  0)  +  cos  a  sin  (360°  -  /8) 

=  sin  a  cos  (—  /3)+  cos  a  sin  (—  /8). 
cos  (a  -  p)  =  cos  a  cos  (360°  -  /3)  -  sin  a  sin  (360°  -  0) 
=  cos  a  cos  (—  /8)  —  sin  a  sin  (—  /?). 


244  UNIFIED   MATHEMATICS 

Substituting     in     these     formulas,     cos  (—  ft)  =  cos  ft,     and 
siu  (—  /3)  =  —  sin  ft  we  obtain  the  subtraction  formulas  : 

sin  («  —  /?)  =  sin  «  cos  ft  —  cos  a  sin  ft, 
cos  (a  —  ft)  =  cos  a  cos  /?  -f-  sin  a  sin  /?. 

7.   Tangent  formulas.  — 


tan  («  +  /?)  =  ^  __ 

1  -  tan  a  tan  /3  1  +  tan  a  tan  ^ 

Since  sin  («  +  /?)  =  sin  «  cos  ft  -f  cos  a  sin  ft, 

and  cos  (a  +  ft)  =  cos  a  cos  ft  —  sin  a  sin  ft, 

for  all  angles  a  and  /3,  without  restriction,  it  follows  that 


tan  («  +  ft)  =  sin  fe  +  £)  =  sin«cosff  +  co3ce  sin/3 
cos  (a  +  /?)      cos  a  cos  /8  —  sin  a  sin  /8  ' 

for  all  angles  a  and  /3. 

Dividing  numerator  and  denominator  of  the  right-hand  ex- 
pression by  cos  a  cos  ft,  we  have 


1  -  tan  «  tan  ft 

o-     M     i  /  tana  —  tan  8 

Similarly,      tan  (a  —  8)  =  — 

l  +  tanatan£ 

8.    Functions  of  double  an  angle.  —  The  formulas  for  sin  (a+ft), 
cos  (a  +  ft),  and  tan  («  -f-  ft}  hold  if  ft  =  a,  whence 

sin  (2  a)  =  sin  («  +  «)  =  sin  a  cos  a  +  cos  a  sin  a  =  2  sin  a  cos  a, 

sin  (2  a)  =  2  sin  a  cos  a. 

Similarly,  cos  (2  a)  =  cos2  a  —  sin2  «  =  2  cos2  a— 1  =  1  —  2  sin2  a. 
By  division  and  simplification,  or  directly  from  tan  («  +  ft), 

2  tan  a 

tan  (2  a)  =  —  —  • 

1  -  tan2  a 

Note  that  whether  a  be  regarded  as  positive  or  negative,  i.e. 
as  obtained  by  positive  or  negative  rotation,  as  +  a,  or 
+  a  —  360°,  2  a  has  the  same  terminal  line  as  2  «  —  720°. 


ADDITION   FORMULAS  245 


NOTE.  —  See  the  preceding  list  of  problems,  and  use  numerical  values 
as  there  computed. 

1.  Given  a  =  45°,  ft  =  30°,  find  sin  («  —  ft)  and  cos  (a  -  ft), 
checking  by  the  tables. 

2.  Given  a  =  60°,  ft  =  45°,  find   sin  (a  —  ft)   and  compare 
with  problem  1.     Find  tan  (a  +  ft),  tan  (a  —  ft),  and  tan  2  a. 

3.  Given  sin  a  =  f  and  sin/2  =-fa,  ^n<^  s^n  (a  ~  P)  when 
«  and  ft  are  acute ;   find  sin  (a  —  ft)  when  a  and  /3  are  both 
obtuse.     Explain  the  result ;  find  sin  («  —  ft)  and  cos  (a  —  ft) 
when     a    is     obtuse    and    ft    is     acute.      Interpret.      Find 
tan  (a  —  ft),  tan  (a  -(-  /?),  and    tan  2  a  for  a  and  /3  in  I. 

4.  Given  a  and  ft  acute,  sin  a  —  .351  and  cos  ft  =  .652,  find 
sin  («  —  ft)  and  check  by  the  tables.     Find  tan  (a  —  ft). 

5.  Given  sin  18°  =  .3090,  cos  18°  =  .9511,  and  sin  15°  from 
problem  1,  find  sin  3°  and  cos  3°. 

6.  Find  sin  42°  as  sin  (60°  -  18°). 

7.  Express  sin  (60°  —  a)  and  cos  (60°  —  a)  in  terms  of  func- 
tions of  a. 

8.  Find  the  value  of  sin  (60°  +  «)—  sin  (60°  —  a). 

9.  Find  the  value  of  cos  (45°  -f  «)+  cos  (45°  —  a). 

10.  Show  that  sin  (a  +  ft)  sin  (a  —  ft)  =  sin2  a  —  sin2  ft. 

11.  Find  a  value  of  cos  (a  +  ft)  cos  (a  —  ft),  similar  to  the 
preceding. 

12.  Given  tan  a  =  1.4,  find  tan  2  a. 

13.  Given  cos  2  a  =  .63,  find  sin  a  and  cos  a ;  are  there  two 
solutions  ? 

14.  Given  that  one  line  cuts  the  x-axis  at  an  angle  a  such 
that  tan  a  =  3,  and  another  line  cuts  the  avaxis  at  an  angle  /3 
such  that  tan  ft  =  \,  find  the  tangent  of  the  angle  between  the 
two   lines   by   assuming   that   they   intersect   on    the   avaxis. 
Check  by  using  the  tables  to  find  the  slope  angles  of  these 
lines. 


246 


UNIFIED  MATHEMATICS 


9.  The  tangent  of  the  angle  between  two  lines.  —  Given  any 
two  lines  as  y  —  3  a;  —  5,  ?/=  —  x  —  7,  it  is  evident  by  plane 
geometry  that  the  angle  between  them  is  the  same  as  the  angle 

between  y  =  3  x,  y  =  —  x, 
lines  parallel  to  these 
given  lines  through  the 
origin.  The  word  "be- 
tween" implies  no  dis- 
tinction as  to  priority  of 
either  line :  thus  the 
angle  may  be  taken  as 
either  a  positive  or  nega- 
tive acute  angle,  or  the 
corresponding  supple- 
mentary angle.  Thus  if 
the  lines  were  inclined 
to  each  other  at  30°, 
the  angle  might  be  con- 
sidered as  -(-  30°,  -  30°, 
+  150°,  or  -150°;  the 
tangent  of  the  angle 
would  then  have  the 


Angle  between  two  lines 

Parallel  lines  through  the  origin  make 

the  same  angle. 

V3 

i 

3 


value  given  by  the  expression  ± 


To  distinguish  between  the  two  lines  we  may  say  that  we 
wish  the  angle  from  the  line  of  slope  +  3  to  the  line  of  slope 
—  1,  or  in  the  general  case,  from 
the  line  of  slope  m2  to  the  line  of 
slope  mi ;  by  analogy  with  our 
use  in  defining  the  angle  which 
a  line  makes  with  the  cc-axis,  when 
we  say  the  angle  which  the  line 
y  =  —  x  makes  with  y  =  3  x  we 

mean  the  angle  obtained   by   re-  Angle  between  two  lines 

volving  the   line   whose   slope  is  <f>  =  el  —  02,  0i  >  &z- 


ADDITION  FORMULAS 


247 


3  so  as  to  make  it  coincide  with  the  line  whose  slope  is  —  1. 
Calling  the  angle  whose  tangent  is  3  (written  tan"1  3  or  arc 
tan  3,  meaning  the  angle  whose  tangent  is  3),  02,  and  the  angle 
whose  tangent  is  —  1,  Ol}  we  find  that  the  angle  <j>  from  the  02 
line  to  the  0j  line  is  <£  =  dl  —  02. 


tan  ^  tan  (ft- ft)  =   tanft-tanft-  =  I 
1  +  tan  Oi  tan  02     1 


If  the  two  lines  are  parallel  the  angle  is  0,  hence  tan  <f>  =  0, 
and  mi  —  w2  =  0,  or  mi  =  w2,  as  anticipated  ;  if  the  lines  are 
perpendicular  tan<£  becomes  infinitely  large,  and  for  finite 
values  of  mt  and  m9_  (excluding  lines  parallel  to  the  axes),  the 

denominator   1  +  m^m^  =  0,   or  m2  =  --  -,  i.e.  the  slope  of  a 


perpendicular  is  the  negative  reciprocal  of  the  slope  of  the 
given  line.  When  one  line  is  parallel  to  the  ?/-axis,  its  slope 
ra2  (or  m{~)  is  infinite,  but  the  angle  between  the  two  lines  can 
be  obtained  by  dividing  numerator  and  denominator  of  tan  <f> 


by   m2  (or  ra^,  giving  tan  <f>  — 


or  --   when  ra2  ap- 


proaches  infinity,  for  the  tangent 
of  the  angle  made  by  a  given  line 
with  the  y-axis. 

tan  4  =  mi-m2    gives  the  an- 

1  -f-  /7Zi/7Z2 

gle  from  the  ra2  line  to  the  mt 
line. 

If   #2  >  Oi,  <f>  is  negative,  but 
the  formula  <£  =  Ot  —  02  still  holds. 


10.   Functions  of  half  an  angle,     cos  (2  a)  =  cos2  a  —  sin2  a  for 
all  values  of  a. 

Substitute  x  for  2  a,  and  ^  for  a. 
2 


248 


UNIFIED  MATHEMATICS 


cos  x  =  cos2 sin2  -• 


l.oos' 


whence 


Half-angle  relations 
Similarly, 


cos  ?  =  ±  V|(l  +  cos  a). 

M 


sin  -  =  ± 
« 


—  cos  cc). 


tan  ft  =  ±     /T-  cos  a;  ^  ±     /(I  -  cos  aQ(l  -  cos  x)  ? 
2  *l  +  cosa;          *  1  — cos2x 

+  if  x  is  in  I  or  II,  and  —  if  a;  is  in  III  or  IV  ;  the  formula 

X        1  —  COS  X   •  •  1-1 

tan  -  = -  is  one  111  which 

2         since 

the  sin  x  takes  care  of  the  algebraic  sign ;  and  so  also 

sin  a; 


x 

tan  -  =  • 


,  both 


2      1  +  cos  x 
by  rationalization. 

Note  that  if  x  is  regarded  as  a  positive  angle,  less  than  360°, 

x 
sin  -    is    + ;  but    the    same  position    of    the    terminal    line 

is    obtained   by    x  ±  360° ;  -'    and    ~  + 180°    have    sine   and 

(x  \  x 

cosine   opposite    in   sign,    but   tan  f  -  ±  180°  ]=  tan  -•     Since 

cos  (—  x)  =  cos  (x~)  it  must  be  stated  whether  x  is  in  I  or  IV,  or 
in  II  or  III ;  i.e.  cos  x  alone  does  not  locate  the  angle  x. 

If  a;  in  I  is  regarded  as  a  positive  angle,  -  is  +  acute,  and 

<u 

/%•  />» 

sin  -  and  cos  -  are  positive  ;  if  x  in  I  is  regarded  as  a  negative 
reflex  angle,  '-  is  negative  obtuse,  and  sin  -  and  cos  —  are  both 

_  a  +i 


ADDITION  FORMULAS  249 


negative ;  in  either  case  tan  -   is  positive.     Similarly  if  x  is 

& 

taken  in  II,  III,  or  IV,  the  formula  takes  care  of  all  positions, 
proper  account  being  taken  of  the  algebraic  sign  of  the  radicals. 


PROBLEMS 

Find  the  angle  between  the  following  lines : 

1.  y  =  3  x  —  5,  and  y  =  —  x  —  7. 

2.  y  =  3  x  —  5,  and  y  =  x  —  7. 

3.  2y-3x-7  =  0,  and  3 y  +  4 x -5  =  0. 

4.  3  y  =  5  x  —  5,  and  y  =  8  x  —  10. 

5.  3  y  =  5  as  —  5,  x  =  5. 

6.  3y  =  5x—  7,  y  =  5. 

7.  In  the  preceding  6  problems,  find  the  tangent  of  the  angle 
made  by  the  first  line  with  the  second  line,  i.e.  the  tangent  of 
the  angle  obtained  by  rotating  the  second  line  until  it  coin- 
cides with  the  first.     Why  is  it  that  the  sense  of  this  rotation 
is  immaterial  ? 

8.  In  the  above  problems  check  by  finding  from  the  tables 
the  trigonometric  angles  involved. 

9.  Find  the  pencil  of  parallel  lines  making  an  angle  of  30° 
with  each  of  the  lines  in  problem  1 ;  find  the  one  of  the  family 
through  (-3,5). 

10.  Find  the  pencil  of  lines  making  an  angle  of  45°  with 
each   of    the    lines    in    problem   3 ;  find    the   particular   one 
through  the  origin. 

11.  Find  the  pencil  of  lines  making  an  angle  of  90°  with 
each  of  the  lines  in  problem  4. 

12.  Given  sin  30°  =  .5000,  cos  30°  =  .8660,  and  tan  30°  =  .5774, 
find  sin  15°,  cos  15°,  and  tan  15°. 

13.  Find  sin7|°,  cos  7-|°,  tan  7^-°,  using  half-angle  formulas. 

14.  Given  sin  45°  =  cos  45°  =  .7071,  find  sin  22£°,  cos  22|°, 
and  tan  22°. 


250  UNIFIED  MATHEMATICS 

15.  Use  sin  (a  —  ft)  formula  to  obtain  sin  7|°  and  cos  7f , 
from  the  functions  of  30°  and  22^-°.    Compare  with  problem  13. 

16.  Given  sin  18°  =  .3090  and  cos  18°  =  .9511,  find  sin  12° 
and  cos  12°. 

17.  From  the  functions  of  12°,  compute  the  functions  of  6°, 
and  then  the  functions  of  3°   and  of   1|°,   using   half-angle 
formulas. 

18.  Compute  sin  If  and  cos  If  by  the  difference  formulas, 
taking  If  as  7f  —  6°. 

19.  Compute  the  functions  of  f °  from  the  functions  of  1°. 

20.  Find  by  interpolation  sin  1°  and  cos  1°  from  the  com- 
puted values  of  the  functions  of  f°  and  If.     Compare  with 
the  tabular  values. 

21.  Make  a  table  of  values  of  the  sine,  from  0  to  45°  in- 
creasing by  If  intervals. 


CHAPTER   XVI 


TRIGONOMETRIC   FORMULAS  FOR   OBLIQUE 
TRIANGLES 

1.  General  statement. — Employing  elementary  theorems  of 
plane  geometry  it  is  possible  to  construct  any  triangle  when 
given  the  three  sides,  or  two  sides  and  an  angle,  or  one  of  the 
three  sides  together  with  two  of  the  angles ;  in  trigonometry 
the  corresponding  problem  is  the  numerical  solution,  not  sim- 
ply the  graphical,  of  the  types  of  triangles  mentioned.  The 
trigonometric  solution  which  has  been  given  of  the  different 
types  of  right  triangles,  with  unknown  parts,  can  be  applied 
to  effect  the  trigonometric  solution  of  any  oblique  triangle; 
but  in  general,  these  methods  do  not  give  convenient  formulas 
for  computation.  As  the  general  triangle  is  fundamental  in 
surveying  (note  the  term  "  triangulation"),  in  astronomical 
work,  and  in  many  problems  in  physics,  more  convenient 
formulas  than  those  given  by  right  triangles  are  a  practical 
necessity. 

In  general  the  laws  and  formulas  of  plane  trigonometry 
connect  directly  with  proposi- 
tions of  plane  geometry ;  the 
effort  is  to  express  the  inter- 
dependence of  the  angles  and 
sides  in  the  form  of  equations 
involving  the  trigonometric  func- 
tions of  the  angles. 

The  vertices  of  any   triangle  A,  B,  C;  a,  p,  -y;  a,  b,  c 

being  lettered  A,  B,  C,  it  is  con- 
venient to  designate  the  corresponding  angles  at  these  ver- 
tices by  a,  ft,  and  y,  respectively,  or  by  A,  B,  and  C,  if  no 

251 


252 


UNIFIED  MATHEMATICS 


confusion  of  meaning  is  possible ;  the  sides  opposite  A,  B,  and 
C  are  designated  by  a,  &,  and  c,  respectively. 

2.   Cosine  law.  —  If  the  two  sides  of  a  triangle  are  given,  the 
third  or  variable  side,  opposite  the  angle  a,  between  the  two 


.HJ 

6'                        \C' 

h 

I. 

A                            \ 

-    -  -  ^s 

.  —  J-^ 

\                                      V 

-/ 

r  —                         ~\ 

\                                 5 

\            ^N 

j 

1  -T/'C  "/ 

3-L((  _i-          _I_J-^.  /  _L_  -  V-^-  - 

\                      _^ 

\                                                \ 

-         ]|-                   t1!                        /V 

--J 

.  ^      L       -                                                        \ 

:/:: 

\                                          \ 

\                                                                  \ 

S                             SV                      ^Sr         - 

ka 

.  —  ^--.         ^a                V 

-l--t            V 

-/ 

:*::  A           \      o       \ 

\-  \    «       \ 

s 

-'/ 

/;     A                 i> 

j/:       .1          as 

a2  =  b2  +  c2  -  2  be  cos  a 

given  sides,  evidently  changes  as  a  changes.  Let  b  and  c 
remain  fixed.  Let  M  be  the  foot  of  the  perpendicular  from 
C  upon  AB ;  then  AM  =  b  cos  a,  for  any  angle  «  when  the 
direction  AB  is  taken  as  positive.  Further  in  every  position 

MB  =  AB-  AM  =  c  -  b  cos  «, 
for  in  every  position  AM  +  MB  =  AB. 
The  altitude  MC  =  h  =  b  sin  a. 
Hence,  SC^  =  JO2  +  MC2' 

—  (c  —  b  cos  a)2  +  (6  sin  «)2 
=  c2  —  2  be  cos  a  +  &2(cos2  a  +  sin2  a). 
a2  =  62  +  c2  -  2  6c  cos  a. 

All  limitations  upon  a  are  removed  by 
the  different  types  of  figures.  Hence 
for  any  angle  a, 

a2  =  b2  +  c2  —  2  &c  cos  a 
gives  the  length  of  the  side  a,  opposite 

a.  in  terms  of  the  other  two  sides  and  a. 
M  falling  outside  B          ~ . 

Formula  unchanged          SmC6  a  and  "    ma^  rePresent   an7   Slde 
and  the  opposite  angle  of  any  given  tri- 
angle, 6  and  c  being  the  other  two  sides,  our  formula  may  be 
stated  as  follows : 


FORMULAS  FOR  OBLIQUE  TRIANGLES         253 


TJie  square  of  any  side  of  a  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  less  twice  their  product  into  the 
cosine  of  the  including  angle. 

Or,  Tfie  cosine  of  any  angle  equals  the  difference  between  the 
sum  of  the  squares  of  the  two  including  sides  and  the  square  of 
the  side  opposite,  divided  by  twice  the  product  of  the  including 
sides. 

If  a,  b,  and  c  are  the  sides  of  any  triangle,  with  a,  ft,  and  y 
the  corresponding  opposite  angles,  we  have  the  following 
relationships  :  a2  =  &2  +  c,  _  2  bc  COs  a, 

'  b2  =  c2  -f  a2  —  2  ac  cos  ft, 
c2  =  a2  +  b2  —  2  ab  cos  y ; 
7,2  _i_  C2  _  a2 


or 


cos  a  = 
cos  ft  = 
cos  y  = 


'2bc 

C2  +  a2  _  ft2 

2ac 


3.  Cyclic  interchange.  —  Any  formula  which  has  been  de- 
rived, without  imposing  any  limitations  upon  a,  b,  c,  a,  ft,  or  y, 
connecting  a,  b,  c,  and  trigono- 
metric functions  of  the  angles  a, 
ft,  and  y,  will  continue  to  hold  if 
a  and  b  and,  at  the  same  time  a 

and  ft,  are  interchanged  ;  or  if  I ( 


are  changed  to 


to       ,  and 


to 


«,      « 

such    changes     effect 


? 


££ 


X 


simply  a  re-lettering  of  the  figure.  Cydic  interchange 

The  change  of  a  into  6,  b  into  c, 

and  c  into  a  is  called  a  cyclic  interchange  of  the  letters  a,  6, 
and  c.  Note  that  cyclic  interchange  gives  the  second  formula 
from  the  first,  and  the  third  from  the  second. 


254  INIFIED  MATHEMATICS 

In  the  figures,  a  in  the  first  is  chosen  as  an  acute  angle,  but 
this  limitation  is  removed  by  deriving  the  same  formulas  for 
a  a  right  angle  and  for  a  an  obtuse  angle ;  c  is  taken  as 
longer  than  b,  but  interchanging  b  and  c  in  our  derived  for- 
mula leaves  the  formula  unchanged ;  assuming  6  and  c  equal 
would  involve  no  change  whatever  in  our  proof;  and  if  b 
is  assumed  greater  than  c,  a  fourth  figure  can  be  drawn  in 
which  M  falls  beyond  B  on  AB  produced ;  but  the  formula 
a2  =  62  -j-  c2  —  2  be  cos  «  remains  the  same,  as  the  student  may 
easily  verify. 

PROBLEMS 

In  the  following  problems  use  .866,  .707,  and  .500  for  the  cosines  of  30°, 
45°,  and  60°,  respectively. 

1.  Given  b  =  140,  c  =  230,  a  =  60°,  compute  a.     Kefer  back 
to  the  section  on  extraction  of  square  root,  page  23. 

2.  Compute  a  when  a  =  30°  and  45°,  90°,  120°,  135°,  180°, 
when  b  =  140,  c  =  230. 

3.  Given   a  =  155,    c  =  234,   ft  =  35°,    compute   b.       What 
changes  in  b  are  effected  by  changes  of  ±  10'  in  ft  ? 

4.  Given  a  =  155,  c  =  234,  compute  ft  when  b  =  172.     What 
is  the  maximum  change  in  ft  which  an  error  of  ±  1  unit  in  a, 
b,  and  c  could  introduce,  ft  being  computed  to  minutes  ?     Take 
1551  234|  with  1711 ;  take  1541,  and  2341,  with  1721     Note 
that   (155i)2  and    (154|)2  differ   from   (155)2  by   about  155; 
similarly  with  the  other  values ;  if  the  squares  are  found  by 
logarithms  it  is  well  to  look  up  log  155.5  and  log  154.5  at  the 
same  time  as  log  155,  etc. 

5.  In  problem  1,  find  cos  ft,  and  then  ft,  taking  for  a  the 
value  obtained  there. 

6.  In  problem  1,  find  cos  y  and  y,  using  the  computed  value 
of  a.     Check  by  summing  ft  and  y  with  the  given  angle. 

7.  Given  a  =  200,  6  =  150,  c  =  300,  find  a.     What  change 
in  a  would  a  change  of  ±  1  in  a  effect  ?     Suppose  that  a,  b, 


FORMULAS  FOR  OBLIQUE  TRIANGLES         255 


and  c  are  given  only  to  two  significant  figures,  i.e.  a  is  between 
195  and  205,  b  is  between  145  and  155,  c  is  between  295  and 
305,  compute  a  and  discuss  limiting  values. 

8.  Compute  the  third  side  in  the  following  5  problems,  using 
logarithms  for  squaring ;  time  yourself  in  the  exercise.  Fifty 
minutes  should  be  ample  time  .for  the  5  problems ;  devise  a 
convenient  form  and  use  it  in  each  example. 

a.  Given  a  =  366,  b  =  677,  y  =  15°  10'. 

b.  Given  a  =  423,  c  =  288,  ft  =  35°  15'. 

c.  Given  b  =  627,  c  =  816,  a  =  100°  41'. 

d.  Given  a  =  635,  c  =  341,  ft  =  67°  38'. 

e.  Given  c  =  184,  b  =  295,  a  =  130°  54'. 

4.  Sine  law.  —  A  circle  may  be  circumscribed  about  any 
triangle  ;  let  the  radius  be  designated  by  R.  The  figure  shows 


Sine  law : 


sin  a      sin 


sin  y 


that  if  A  is  an  acute  angle,  sin  «  =  -2-?  =  JL ;  if  «  is  90°,  this 

Jf      2  R 

formula  is  still  true,  as  a  equals  2  R,  and  the  formula  gives 

sin  90°  =  1 ;  if  « is  obtuse,  the  figure  gives  sin  (180°  —  a)  =  -^-, 

a 

whence  sin  a  = . 

2R 


256 


UNIFIED  MATHEMATICS 


Therefore  without  any  limitation  whatever  upon  a, 


sin  a  = 


2R' 

interchange  of  letters  gives 


sn  ?  = 


sin  y  =  7 


_ 
2R' 

c 


'2R 


Whence 
a 


Sine  law :    a.  obtuse 


sin  a  sin  B  sin  y 
This  formula  states  that  in  any 
triangle  the  ratio  of  the  side  op- 
posite any  angle  to  the  sine  of  that  angle  is  constant,  and  this 
ratio  is  numerically  equal  to  the  diameter  of  the  'circumscribed 
circle. 

-n,     ,,        sin  a      sin  8      sin  y  , .      c  ,, 

£  urther,  -    —  =  —  £-  = L  }  Or  the  ratio  of  the  sine  of  any 

a  b  c 

angle  to  the  side  opposite  is  constant. 

Note  that  if  2  R  is  regarded  as  the  chord  of  180°  of  the 
circle  in  which  the  triangle  ABC  is  inscribed,  the  proposition 
states,  in  effect,  that  in  any  circle  any  chord  is  proportional 
to  the  sine  of  the  inscribed  angle  which  intercepts  the  arc  of 
the  chord. 

The  formula  may  be  stated : 

sin  a  _  sin  B  _  sin  y  _  sin  90°  _  sin  30°  _        sin  k° 
a  b  c  2R  R      ~  chord 


all  of  the  chords  being  chords  of  the  circle  circumscribed  about 
the  triangle.  The  ratio  of  the  sine  of  any  central  angle  in  a 
circle  to  the  chord  of  double  the  angle  can  readily  be  shown  to 

be  constant,  — —  • 

5.   The  sine  law  historically.  —  The  sine  law  was  discovered 
by  an  Arabic  (Persian,  by  birth)  mathematician,  Nasir  al-Din, 


FORMULAS  FOR  OBLIQUE  TRIANGLES  "      257 

at-Tusi,  who  lived  1201-1274  A.D.  To  him  we  owe  the  first 
systematic  treatise  on  plane  trigonometry,  an  achievement  made 
possible  by  the  combination  of  the  Greek  trigonometry  using 
chords  with  the  Hindu  trigonometry  employing  sines.  To 
Europeans  the  sine  law  was  communicated  by  the  great  Ger- 
man mathematician  and  astronomer,  Regiomontanus,  in  his 
work  on  trigonometry,  De  Triangulis,  the  first  published  sys- 
tematic treatise  ;  it  was  published  at  Nuremberg  in  1539,  many 
years  after  the  death  of  Regiomoutanus,  who  lived  1436-1476. 

PROBLEMS 

1.  Given  a  =  150,  b  =  200,  a  =  30°,  find  sin  (3  using  natural 
functions. 

2.  Given  a  =  150,  a  =  30°,   ft  =  45°,   find  6,   using  natural 
functions. 

3.  Given  a  =  150.4,  b  =  214.3,  a  =  31°  10',  find  sin  ft  employ- 
ing logarithms. 

4.  Given  o  =  150.4,  a  =  31°  10',  ft  =  44°  16',  find  b  by  loga- 
rithmic computation. 

5.  In   the   formula,   a2  =  b2  +  c2  —  2  be  cos  a,  substitute  the 
values  as  given  in  problem  1  and  solve  for  c.     Note  that  there 
are  two  solutions.     What  is  the  explanation  ? 

6.  Time  yourself  in  solving  the  following  set  of  6  problems, 
applying  the  sine  law ;  make  a  type  form  of  solution  and  use 
it  in  each  problem.     Thirty  minutes  should  be  sufficient  for 
the  6  problems. 

a.  Given  a  =  366,  6  =  677,  a  =  15°  10'.     Find  sin  ft  and  ft. 

6.  Given  a  =  423,  c  =  288,  y  =  35°  15'.     Find  sin  a  and  a. 

c.  Given  a  =  627,  a  =  100°  11',  ft  =  43°  15'.     Find  6. 

d.  Given  6  =  816,  ft  =    67°  18',  y  =  34°  9'.       Find  c. 

e.  Given  c  =  635,  ft  =  130°  14',  a  =  20°  12'.     Find  b. 
/.  Given  b  =  284,  a  =   40°  10',  ft  =  35°  15'.     Find  c. 


258 


UNIFIED  MATHEMATICS 


6.  Half-angle  formulas.  —  As  the  circumscribed  circle  has 
yielded  a  formula  of  great  value  trigonometrically  the  in- 
scribed circle  may  be  examined  trigonometrically  with  the  hope 

of  a  similar  result. 

The  bisectors  of 
the  three  angles  of 
the  triangle  meet  in 
a  point  which  is  the 
center  of  the  in- 
scribed circle ;  let 
this  circle  be  drawn 
and  let  L,  M,  N,  be 
the  points  of  tan- 
gency,  then 

AM=AN, 
BL  =  BN, 

OL  =  OM  =  ON  =  r,  radius  inscribed  circle  CL  —  CM, 

being  tangents  from  an  exterior  point.  Evidently  the  six 
segments  mentioned  make  the  perimeter,  2  s,  of  our  triangle ; 
2s  =  a  +  6-f-c;  adding  above  we  have  that 

AM+  BL  +  CL  =  AN+  BN+  CM=  s, 
but  BL+CL  =  a,  and  BN+  AN=  c,; 
whence  AM  =  s  —  a,  CM=  s  —  c,  and  similarly 
BN=BL  =  s-b 
ON=  OL  =  OM=  r. 


tan  _  «  =  • 


s—  a 


a.  -1-    /->  ' 

tan  ~B  =  —  —'. 
2^      s-b 


2 


s  —  c 


7.  Area.  —  In  the  preceding  section  the  area  of  the  given 
triangle  is  easily  determined  in  terms  of  r  and  s,  for  the  area 
equals  the  sum  of  the  three  triangles  on  a,  b,  and  c  as  bases, 
each  having  the  altitude  r.  .•.  A  =  %  r(a  +  b  +  c)  =  rs. 


FORMULAS  FOR  OBLIQUE  TRIANGLES         259 

However,  if  the  three  sides  are  given,  this  formula  does  not 
enable  us  to  determine  r  without  using  a  further  formula  to 
determine  A, 

A  =  ^bc  sin  a  =  ^  ac  sin  ft  =  1  ab  sin  y. 

This  type  of  formula  for  the  area  is  applicable  when  two  sides 
and  the  included  angle  of  a  triangle  are  given  or  found. 

&2  4-  c2  —  a2 

A=  4-  6c  sin  a  can  be  combined  with  cos  «=  —  -£-         -  in  such 

2  be 

a  way  as  to  eliminate  a,  giving  A  in  terms  of  the  three  sides. 
A*  =  $  &2c2  sin2  a  =  £  &2c2  (1  -  cos2  a)  . 

=  £  &2c2  (1  -  cos  a)(l  -f  cos  a) 

c2  -  qg\          62  +  c2  -  a 


4  V  2&c 

Wa«  -  62  +  2  6c  -  c2\/62+  2  6c  +  c2  -  a2' 


4  V  26c  2bc 

c)2  _  ag] 


16  62c2 

_  (a—  6  +  c)(«  +  ft  —  c)(6  +  c  —  a)  (6  +  c  +  a). 
2        •       2        •        2  2 


But 


The  above  formula  for  A2  may  be  written, 
A2  =  s(s  —  a)(s  —  &)(*  —  c) 
A  =  Vs(s  —  d)(s  —  6)(s  —  c). 


-,     ,,  A 

Further  A  =  rs,  whence  r  =  —  = 

s 

This  value  of  r  is  employed  with  the  half-angle  tangent 
formulas  of  the  preceding  article  to  determine  the  angles  of  a 
triangle  when  the  three  sides  are  given. 


260 


UNIFIED  MATHEMATICS 


8.  Newton's  check  formula.  —  A  formula  which  involves  all 
of  the  sides  and  all  of  the  angles  of  an  oblique  triangle  is 
particularly  desirable  as  a  check  formula  to  be  used  upon  the 
results  obtained  by  direct  application  of  the  sine  law  or  in  a  solu- 
tion obtained  by  right  triangles.  Such  a  formula  was  devised 
by  Sir  Isaac  Newton  and  appeared  in  his  Arithmetica 
of  1707 ;  our  proof  follows  the  lines  of  that  by  Newton. 


B 


3rf 


2 


?j£ 

s  .At 


•P 


.1 


F 


Let  ABC  be  any  triangle ; 

from  C  draw  the  bisector  CE  of  the  angle  -4(7J5  or  y ; 

extend  5(7  to  F,  making  CF=CA=b; 

AF  is  parallel  to  CE,  by  plane  geometry ; 

angle  CFA  =  angle  BCE  =  i  y. 
Now  angle  .BvlF=  «  +  i  y  =  90°  - 1(«  -  £), 
since  !a  +  !/?  +  ^y=  90°. 

Applying  the  sine  law  to  the  triangle  BAF,  we  have  the 
desired  formula : 

a  +  b cos  ^(g  —  (3) 

c  sin  \  y 


FORMULAS  FOR  OBLIQUE  TRIANGLES         261 


By  drawing  the  bisector  of  the  exterior  angle,  and  drawing 
a  parallel  from  A,  a  second  useful  check  formula  is  obtained  : 
a  —  b  __  sin  |(q  —  P) 

c  cos  i  -y 

Noting   that   1  y  =  90°  —  ^(a  +  /?),   division   of  the   second 
equation  by  the  Newtonian,  member  for  member,  gives 


a  +  b     ten  £  (a  +  p) 
this  symmetrical  formula  is  known  as  the  tangent  law. 

Cyclical  interchange  gives  in  each  one  of  the  above  two 
corresponding  formulas. 

9.  Historical  note.  —  The  formula  A  =  Vs(s  —  d)(s  —  b)(s  —  c) 
was  first  given  by  Hero  of  Alexandria,  first  century  A.D., 
a  teacher  of  mathematics  and  mechanics  in  what  was  probably 
a  kind  of  technical  school  at  Alexandria  in  Egypt  ;  it  is  called 
Hero's  formula, 

An  extension  of  this  formula  is  given  by  Bhaskara,  a  Hindu 
mathematician  of  about  1000  A.D.  Bhaskara's  formula  gives 
the  area  of  any  quadrilateral  which  is  inscribable  in  a  circle, 
i.e.  with  the  opposite  angles  supplementary,  as 


L- 


A  =  V(s  —  a)(s  —  b)(s  —  c)(s  —  d). 
The  triangle  may  be  regarded  as  a  special  case  with  d  =  0. 

10.   Reflection  and  refraction  of  light.  —  Rays  of  light,  like  rays 
of  heat  and  sound  and  elec- 
tric rays  of  various  types, 
travel  in  straight  lines  from 
the  source.     Rays  of  light 
emanating    from    the    sun 
travel    in    nearly    parallel 
rays,  since  the  point  of  con- 
vergence, the  source  at  the 
sun,    is  at    SO    great  a  dis-         Reflected  ray  travels  shortest  path 
tance  from  the  earth.     A  ray  of  light  which  meets  a  polished 
plane  surface,  a  mirror,  is  reflected  at  an  angle  which  is  such 


262 


UNIFIED  MATHEMATICS 


as  to  make  the  total  path  from  the  source  (L)  to  the  reflect- 
ing surface  and  then  to  a  second  position  (R)  the  shortest 
possible.  LSR  is  the  shortest  distance  from  L  to  S  to  R  if 
the  angle  of  incidence  i,  made  by  the  original  ray  with  the 
normal,  to  the  surface  at  S  where  the  ray  strikes,  is  equal  to 
the  angle  of  reflection  r.  Evidently  LSR  =  LSR' ;  the  straight 
line  joining  L  to  R',  a  point  symmetrically  situated  to  R  with 
respect  to  the  polished  surface,  is  shorter  than  any  other  line, 
for  any  other  broken  line  LS'R  =  LS'R'  is  greater  than  the 
straight  line  LSR'  and  hence  greater  than  LSR. 

If  the  ray  of  light  meets,  not  a  polished  surface  but  some 
transparent  medium,  other  than  that  in  which  the  ray  is 
traveling,  the  ray  of  light  is  not  continued  in  the  same 
straight  line  in  which  it  starts  but  it  is  broken,  or  refracted, 
continuing  on  its  path  in  a  straight  line  which  makes  a  differ- 
ent angle  with  the  normal  than  does  the  original,  incident  ray. 

It  is  found  by  physical  ex- 
periments that  the  angle  of 
refraction,  the  angle  of  the 
refracted  ray  with  the  nor- 
mal, bears  a  simple  relation 
to  the  angle  of  incidence, 


sin  i 
sinr 


=  Jc,  wherein  k  depends 


Refracted  ray  of  light 


upon  the  nature  of  the  two 
media  through  which   the 

light  is  passing.     Thus  for  a  ray  of  light  passing  from  air,  a 
rarer  light  medium,  to  the  denser  water  the  value  of  k  is  |, 
sin  i  _  4 
sinr     3 

A  student  who  thoughtfully  examines  this  formula  will  be 
reminded  of  the  sine  law,  which  does  indeed  give  a  very  simple 
construction  for  the  refracted  ray  when  the  constant  k  is  known. 

Let  two  concentric  circles  be  drawn  whose  radii  bear  to 
each  other  the  ratio,  •£,  of  the  index  of  refraction.  In  the 


FORMULAS  FOR  OBLIQUE  TRIANGLES         263 

figure  the  ratio  is  taken  |,  the  index  of  refraction  for  light 
from  air  to  water.  Extend  LO,  the  incident  ray,  to  L',  cutting 
the  circle  of  smaller  radius.  From  L'  drop  a  line  parallel  to 
the  normal  N'O  to  cut  the  larger  circle  in  R.  Connecting  R 


LO  is  the  incident  ray ;  OR  is  the  refracted  ray 

with  0  gives  the  refracted  ray.  In  the  triangle  OL'R,  the 
Z  OL'R  =  180°  - 1,  and  the  Z  ORL'  =  /  r,  of  refraction ;  by 
the  sine  law 

sin  (180°  -  Q  _  sin  i  _  4 
sin  r  sin  r     3 

From  water  to  air  the  index  is  f ,  it  being  found  that  if  the 
refracted  ray  is  replaced  by  an  original  ray,  this  new  ray 
in  the  second  medium  will  be  refracted  along  OL,  the  path  of 
the  incident  ray  with  which  we  started. 

The  construction  for  the  refracted  ray  in  air  for  a  ray  of 
light  emanating  from  the  water,  RO,  is  entirely  similar  to  the 
preceding.  RO  is  extended  to  R1  on  the  larger  circle.  From 


264 


UNIFIED  MATHEMATICS 


R'  a  parallel  R'L  is  drawn  to  the  normal  to  cut  the  smaller 

circle.     OL  is  the  refracted  ray. 

Evidently  if  sin  r  =  f  sin  i,  sin  r  is 
always  less  than  f.  If  a  ray  of  light 
starts  from  any  point  within  the  arc 
A' T  wherein  T  is  the  intersection  of 
the  vertical  tangent  to  the  smaller 
circle  with  the  larger  circle  it  cannot 
be  refracted  into  the  air  at  O,  and  the 
whole  light  is  reflected  at  0.  This 
property  of  the  light  rays  is  utilized 
in  certain  spectroscopic  work.  Thus  in 
the  case  of  a  glass  prism,  index  of  re- 


Glass     reversing    prism 
Index  of  refraction  f 

Angle  of  incidence,45°.    fraction  f ,  if  light  strikes  the  plane  sur- 
For  any  angle  i  greater    face  at  an  angle  of  incidence  greater  than 

41°  48',  since  sin  41°  48'  =  .6666,  or  |, 
all  the  light  will  be  reflected ;  this  type 
of  prism  is  used  in  projecting  lanterns. 


than  a,  sin  a  =  f ,   the 
beam  is  reflected. 


PROBLEMS 

1.  Given  a  —  9,  b  =  14,  c  =  19,  find  the  area  of  the  triangle, 
using  Hero's  formula. 

2.  Given  a  =  9,  b  =  14,  c  =  19,  find  a,  using  the  cosine  law,. 

3.  Given  a  =  9,  b  =  14,  c  =  19,  find  the  area  by  the  formula 
A  =  %bc  sin  a. 

4.  Given  a  =  9.34,   &  =  14.31,  c  =  19.27,   find  the  area  by 
Hero's  fcrmula,  using  logarithms. 

5.  Given  a  =  9.34,  b  —  14.31,  c  =  19.27,  find  a  by  the  cosine 
law,  and  then  find  the  area  using  the  formula  involving  sin  a. 

6.  In  the  two  triangles  above  find  r,  the  radius  of  the  in- 
scribed circle,  using  r  •  s  =  A. 

7.  In  the  two  angles  above  find  tan  i  «,  tan  ^  ft,  and  tan  ^  y, 
using  the  half-angle  formulas.     Find  the  angle  sum  in  each 
case. 


FORMULAS  FOR  OBLIQUE  TRIANGLES          265 

8.  Draw  circles  with  radii  two  inches  and  three  inches  and 
show  how  to  construct  the  refracted  rays  of  light  passing  from 
air  into  glass  at  angles  of  incidence  of  30°,  45°,  and  60°. 

9.  For  what  angle  will  a  ray  of  light  passing  from  glass 
into  water  be  reflected,  and  not  refracted  ?     The  index  of  re- 
fraction of  light  passing  from  glass  into  water  is  -|.     Draw  the 
figure. 

10.  Find  the  angle  of  refraction  of  rays  of  light  passing  from 
air  into  water,  Tc  =  1 .33,   when   the   angles   of  incidence  are 
31°  15',  37°  18',  44°  25',  67°  10',  83°  15'.     For  which  of  these 
angles  is  the  course  of  the  ray  changed  by  the  greatest  amount  ? 

11.  Suppose  the  rays  in  problem  10  to  pass  from  air  into 
glass,  solve  for  the  angles  of  refraction. 

12.  Construct  two  of  the  figures  in  both  problems  10  and 
11,  and  check  graphically  the  results  obtained  above. 


CHAPTER    XVII 

SOLUTION   OF  TRIANGLES 

1.  Solution  of  triangles  given  two  angles  and  one  side:  cap 
type.  —  With  surveying  instruments  the  simplest  method  of 
locating  the  distances  from  two  fixed  points  to  a  third  inacces- 
sible point  is  to  determine  the  length  AB  and  the  angles  a  and 
ft,  at  A  and  B  respectively,  wherein  A  and  B  are  two  points 
from  which  (7  is  visible.  Using  the  sine  law, 


sin  a      sin  ft      sin  y 

we  select  the  equation  -  =  -  , 
sm  a      sin  y 

b  c 

or—   —  =  --  -,    since  in  each   of   these   only   one   unknown 
sin  ft      sin  y 

quantity  appears.     The  third  equation  -  =  -  ,  not   in- 
,  sin  a      sin  ft 

dependent  of  the  other  two,  is  used  as  a  partial  check  upon 
the  computed  values.  As  a  more  complete  check  use  Newton's 
formula, 


cos     a  - 


c  sin  ^  y 

This  form  of  triangle  appears  in  the  classical  problem,  whose 
solution  by  plane  geometry  is  ascribed  to  one  of  the  seven  wise 
men  of  Greece,  Thales  of  Miletus,  sixth  century  B.C.  The  prob- 
lem is  familiar  to  the  surveyors,  being  used  in  determining 
distances  across  a  stream,  or  to  an  inaccessible  point.  The 
astronomer  has  the  same  problem  in  locating  the  distance  of 
fixed  stars  using  two  observations,  at  different  points  in  the 
earth's  orbit,  of  the  angle  made  by  lines  from  the  earth  to  the 

266 


SOLUTION  OF  TRIANGLES 


267 


star  and  to  the  sun  ;  for  simplicity,  the  two  points  of  observa- 
tion may  be  considered  as  taken  at  the  extremities  of  the 
diameter  of  the  earth's  path. 

In  locating  •  batteries  by  the  sound  waves  this  type  of 
triangle  is  employed ;  two  or  three  observers  at  different 
points  can  locate  an  enemy  battery  by  this  method  within  a 
radius  of  fifty  feet  or  thereabouts. 

2.  Type  form  of  solution  :  cap  type.  —  The  form  of  the  solu- 
tion is  important ;  follow  the  given  form  closely. 

Given  a  =  65°  11',  ft  =  38°  24',  c  =  175  feet.     Find  a  and  b. 

c  sm  (t  /      •  1 1        r          i_i       /»    .      i        o/  c 


(written  from  the  formula = 

sin  y  sm  a      sin  y 

should  not  be  set  down). 


which 


sin  y 


sin  a 


,      • 
=  -bc  sma= 


1  c2  sin  a  sin 


logc  = 

+  log  sin  a  = 


2        sin  y 

a  =  65°  11' 
B  =  38°  24' 
y  =  76°  25' 

=    2.2430 

9.9580  -  10 


12.2010  -  10 

-log  sin  y=    9.9876  —  10 
log  a  =    2.2134 

a  =    163.4 

log  a  =    2.2134 

+  log  sin  ft  =    9.7932  -  10 
12.0066  -  10 

-log  sin  a  =    9.9580  -  10 
log  b  =    2.0486  -  10 

But  log  b  =  2.0486  by  above 
computation,  which  checks. 


Two  angles 

logc 

+  log  sin  B 

-  log  sin  y 
log  b 
b 

logc2 

+  log  sin  a 
+  log  sin  B 

-log  sin  y 
log  2  A 

2A 
A 


and  a  side  given 

=    2.2430 
=    9.7932  -  10 
12.0362  -  10 
=    9.9876  -  10 
=    2.0486 
=    111.8 
=    4.4860 
=    9.9580  -  10 
=    9.7932  -  10 
14.2372  -  10. 
9.9876  -  10 
4.2496 
17,760 
8880 


268  UNIFIED  MATHEMATICS 

The  check  which  we  have  used  is  only  partial  as  an  error  in 
y  or  siiiy  would  be  carried  through  the  work  without  showing 
up  in  the  check.  The  Newtonian  formula  gives  a  real  check 
upon  the  computation. 

Check.  q±& 


c  sin  i  y 

a  +  b  =  275.2      log  (a  +  b)          =  2.4396 

log  c  =  2.2430 

.1966 

a  _  £  =  26°  47'     log  cos  |(a  -  ft)  =  9.9880 
y  =  76°  25'     log  sin  £  y  =  9.7914 

.1966  which  checks. 

NOTES.  —  The  whole  form  of  solution  is  placed  on  paper  before  the 
logarithms  are  inserted.  Place  the  given  angles  in  vertical  column  and 
obtain  the  third  angle  by  noting  the  angle  which  added  to  the  given 
angles  makes  180°  ;  thus,  here  note  first  that  to  complete  11'  and  24'  to 
1°  takes  25'  .  Add  this  1°  to  the  8°  and  5°,  the  units  of  our  given  angles, 
making  14°  ;  complete  by  6°,  which  is  written  in  its  proper  place,  to  20°. 
Carry  the  2  tens,  to  the  tens,  making  11  tens,  or  110°,  requiring  7  tens 
(written  in  the  proper  place)  to  complete  to  180°. 

Look  up  logc,  i.e.,  log  175,  writing  this  immediately  in  all  places 
where  it  occurs  ;  for  the  area,  it  is  simpler  to  calculate  2  A  and  divide  by 
two  than  to  divide  by  subtracting  log  2  in  the  work,  log  c2  =  2  log  c, 
which  is  set  down  in  its  place.  Finish,  as  far  as  possible,  with  the  logs 
of  numbers  before  taking  up  the  logs  of  trigonometric  functions. 
log  sin  65°  11',  log  sin  38°  24',  and  log  sin  76°  25'  should  be  found  in  the 
order  in  which  they  occur  in  the  tables,  to  avoid  useless  thumbing  back 
and  forth  ;  any  value  found  should  be  immediately  inserted  wherever 
it  occurs  in  the  form. 

PROBLEMS 

1.  Prove  the  sine  law  by  using  perpendiculars  dropped  from 
a  vertex  to  the  opposite  side. 

2.  Given  c  =  350.4,  a  =  36°  14',  ft  =  100°  24',  find  b  and  a, 
by  the  sine  law. 

3.  Given  a  =  .03504,  a  =  36°  14',  ft  =  100°  24',  find  b  and  c, 
by  the  sine  law. 


SOLUTION  OF  TRIANGLES 


269 


4.  Solve  completely  the  following  5  triangles ;  take  the 
time  of  your  solutions ;  write  the  complete  form  of  solution 
for  each  problem,  in  turn,  before  inserting  any  logarithms. 
The  five  problems  should  be  completed  within  one  hour  and 
20  minutes  using  the  rough  check  by  the  sine  law.  As  a 
separate  exercise  check  all  by  Newton's  formula,  timing 
yourself. 

a.  a  =  627  a  =  100°  11'  /8  =  43°  15' 

6.   6  =  816  0  =  67°  18'  y  =  34°09' 

c.  c  =  635  £=130°  14'  a  =  20°  12' 

d.  &  =  284  a  =  40°  10'  £  =  35°  15' 

e.  a  =  366  a  =  15°  10'  £  =  95°  14' 

3.  Given  two  sides  and  the  angle  opposite  one:  aba.  type. — 
Given  b,  a,  and  a  to  construct  the  triangle  geometrically.  AC 
is  laid  off  of  length  b  and  the  line  AX  is  drawn  so  as  to  make 
Z.  CAX  =  a.  Since  a  must  lie  opposite  to  «,  a  is  taken  as 


Given  two  sides  and  the  angle  opposite  one 

The  side  opposite  the  given  angle  must  always  be  greater  than,  or  equal 
to,  the  corresponding  altitude. 

radius  and  with  C  as  center  an  arc  is  swung  to  cut  the  side 
AX.  Since  the  shortest  distance  from  (7  to  AX  is  the  length 
of  the  perpendicular  CM,  if  a  is  given  less  than  this  perpen- 
dicular there  is  no  solution.  If  a  is  given  equal  to  the  per- 
pendicular there  is  one  solution;  if  a  is  greater  than  the 
perpendicular  the  arc  cuts  AX  in  two  points,  but  unless  a  <  6 
the  one  point  of  intersection  to  the  left  of  A  will  not  repre- 
sent a  solution.  The  perpendicular  is  of  length  b  sin  a ;  if  a 


270  UNIFIED  MATHEMATICS 

is  equal  to  or  greater  than  90°,  there  will  be  one  solution  if 
a  >  b,  and  none  if  a  j<  b,  for  the  greater  angle  lies  opposite 
the  greater  side.  By  plane  geometry  then,  we  have  the  fol- 
lowing scheme,  indicating  whether  one  solution,  two  solutions, 
or  no  solutions  are  possible. 

a  J>  90°,  a  <_  6,  no  solution. 

a  _>  90°,  a  >  b,  one  solution. 

a  <  90°,  a  <  6  sin  a,  no  solution. 

a  <  90°,  a  =  b  sin  «,  one  solution. 

a  <  90°,  b  sin  a  <  a  <  b,  two  solutions. 

a  <  90°,  a  >  b,  one  solution. 

Trigonometrically,  by  our  formulas,  we  would  arrive  at 
these  facts,  but  a  student  who  is  not  able  to  observe  the 
geometrical  relationships  is  not  likely  to  be  able  to  interpret 
the  trigonometric  formulas.  When  the  sine  of  an  angle  is 
given,  the  angle  may  be  either  in  I  or  II,  a  or  180°  —  a  if  a  is 
either  angle  which  satisfies  the  relationship.  Then, 


if  a  <  b  sin  a,  sin  ft  will  be  greater  than  1  and  there  is  no 
angle  satisfying  the  relationship  ;  if  a  >  b,  a  >  ft  (greater 
angle,  greater  side  opposite),  and  only  the  acute  angle  ft  can 
be  taken  ;  if  a  <  b,  both  values  of  ft  can  be  taken. 

PROBLEMS 

1.  Given  a  =  30°,  a  =  150,  b  =  60,  70,  75,  100,  150,  180,  and 
200  respectively  ;  draw  the  figures  and  determine  the  number 
of  solutions  in  each  case.     Solve  for  ft  in  each  case  where  it 
is  possible. 

2.  Given  a  =  90°,  a  =  150,  b  =  75,  100,  150,  200.     Discuss. 

3.  Given  a  =  150,  b  =  75  ;   a  =  20°,  30°,  45°,  60°,  80°,  90°, 
120°,  150°.     Discuss  the  solutions,  geometrically  and  trigouo- 
metrically. 


SOLUTION  OF  TRIANGLES  271 

4.  Solve  the  following  eight  problems,  having  one  or  two 
solutions,  and  time  yourself.  Use  the  following  form  of  solu- 
tion. The  eight  problems  should  be  completed  within  one 
hour. 

a.   Given  a  =  366,  b  =  677,  «  =  15°  10'  ;  solve  for  ft. 


a 

log  6  = 
+  log  sin  a  = 


—  log  a  = 
log  sin  /3  = 


or  simply  ft  =  ,  if  there  is  only  one  solution. 

6.  Given  a  =  423,     c  =  288,     y  =  35°  15'  ;  find  a. 

c.  Given  b  =  376,     c  =  804,     y  =  68°  20'  ;  find  (3. 

d.  Given  b  =  650,    a  =  830,     «  =  98°  56'  ;  find  ft. 

e.  Given  a  =  67.2,    c  =  40.4,    y  =  24°  49'  ;  find  a. 
/.  Given  b  =  .0188,  c  =  .0196,  y  =  100°  14'  ;  find  ft. 
g.  Given  a  =  504.2,  c  =  1763,  «  =  12°  39'  ;  find  y. 

h.   Given  b  =  3,245,000,  c  =  2,488,000,  ft  =  80°  28'  ;  find  y. 

4.    Type  form  :  aba  type  with  two  solutions.  — 

Form  of  solution  when  two  solutions  are  found. 
Given  a  =  187,  b  =  235,    a  =  37°  15'. 


a  sin  a 

Or  Newton's  check  formula, 

b  +  a  _  cos  ^  (ft  —  a) 

c  sin  £  7 

log  6=    2.3711 
+  log  sin  a  =    9.7820  —  10 
12.1531  -10 
—  log  a  =    2.2718 
log  sin  0  =    9.8813  —  10 


272 


UNIFIED  MATHEMATICS 


ft  =  49°  83' 
a  =  37°  15' 
.71  =  93°  12' 
loga=    2.2718 
+  log  sin  7!  =    9.9993  —  10 
12.2711  -  10 

—  log  sin  a  =   9.7820—  10 

log  ci  =    2.4891 
Ci  =    308.4 
log&  =    2.3711 
+  log  sin  7!  =    9.9993  -  10 
12.3704-10 

—  log  sin  ft  =    9.8813  —  10 

logd  =    2.4891 

Compare  with  values  for  log  c\  (and  lo 

2  AI  —  ab  sin  71 
log  a  =  2.2718 
+  log  b  =  2.3711 
4-  log  sin  71  =  9.9993  —  10 
log  2^  =  4. 6422 
2  AI  =  43870 
AI  =  21935 


02  =  130°  27' 
a  =  37°  15' 
72  =  12°  18' 
loga=    2.2718 
+  log  sin  72  =    9.3284  —  10 
11.6002-  10 

—  log  sin  a  =    9.7820  -  10 
logC2=    1.8182 

c2  =  65.8 
log  6=    2.3711 
+  log  sin  72  =    9.3284  —  10 
11.6995-10 

—  log  sin  02  =    9.8813  —  10 
logC2=    1.8182 

c2)  found  above. 

2  A2  =  ab  sin  72 
log  a  =  2. 27 18 
log  b  =  2.3711 
log  sin  72  =  9.3284  -  10 
log  2  A2  =  3.9713 
2  A2  =  9360 
A2  =  4680 


5.  Given  two  sides  and  the  included  angle :  type  airy.  —  The 
method  of  solution  here  given  is  the  solution  by  right  triangles, 
since  that  involves  no  new  formula  and  no  greater  amount  of 
computation  than  the  common  solution  employing  a  new  tan- 
gent formula ;  the  tangent  formula  is  given  in  Section  8  of  the 

preceding  chapter. 

Solution  by  right  triangles. 


m 


Given  a  =  280, 
b  =  240, 


7  =  35°, 


h  =  a  sin  35°, 
x  =  a  cos  35°, 
z  =  b  —  a  cos  35°. 


asin35° 


;c=asin>;  check,  c  =  &  sin  ^ 
z      b  —  a  cos  35°  sin  a  sin  /3 


SOLUTION  OF  TRIANGLES 


273 


2  A 

log  a  • 
+  log  sin  35°  : 

log  h  -. 

-  log  Z   : 

log  tan  a  -. 

a  •. 

?'• 

log  a  : 
-f-  log  sin  7  : 

—  log  sin  a  •. 

logC  : 
C  : 

log  b  : 

+  log  a : 
+  log  sin  7 

log  2  J.  : 

2  A 

A 


-.  ab  sin  7 

2.4472 

9.7586  -  10 

2.2058 
: 1.0253 

1.1805 

:  86°  13' 

:  58°  47' 
;  2.4472 
:  9.7586 -10 

2.2058  -  10 
;  9.9990  -  10 
;  2.2068 

161.0 

2.3802 
:  2.4472 
:  9.7586  -  10 
:  4.5860 
: 38550 
= 19270 


log  a  =  2.4472 
+  log  cos  35°  =  9.9134  -  10 
log  x  =  2.3606 
x  =  229.4 
z  =  10.6 


Check.       log  6  =    2.3802 

+  log  sin  7  =    9.7586 

12.1388 

-logsin/3=    9.9321 
logc=    2.2067 


This  problem  also  occurs  frequently  in  surveying.  It 
permits  the  determination  of  the  direction  and  length  of  a 
tunnel  through  a  mountain  by  means  of  the  location  of  some 
point  from  which  both  ends  of  the 
proposed  tunnel  are  visible.  The  dis- 
tance and  direction  from  a  given  point 
to  a  second  point,  past  some  barrier, 
are  determined  by  this  method.  Thus 
the  distance  from  B  to  C  through 
woods  can  be  found,  if  some  point  can 
be  located  from  which  both  C  and  B 
are  visible.  The  distance  and  direc- 
tion to  invisible  points  are  constantly 
needed  in  artillery  fire ;  another  method  of  finding  distance 
and  direction  of  the  target  is  that  of  finding  an  observation 
point  from  which  both  the  gun  and  the  target,  invisible  at 
the  gun,  are  visible  to  an  observer. 


Distance  across  a  barrier 


274  UNIFIED  MATHEMATICS 

PROBLEMS 

Using  the  form  of  solution  above,  solve  for  the  side  opposite  the  given 
angle. 

1.  Given  a  =  3846,      &  =  4977,      y  =  38°10'.       Find  c. 

2.  Given  b  =  4.832,     c  =  8.973,     a  =  108°  56'.     Find  a. 

3.  Given  a  =  .0485,     c  =  .0682,     (3  =  58°  38'.       Find  6. 

4.  Using  the   form  of  solution  given  above,  find  the  side 
opposite  the  given  angle  in  the  following  five  problems ;  time 
yourself,  and  compare  with  the  time  for  the  same  five  problems 
solved  by  the  cosine  law  (page  255). 

a.  Given  a  =  366,  6  =  677,  y  =  15°10'. 

b.  Given  a  =  423,  c  =  288,  ft  =  35°  15'. 

c.  Given  6  =  627,  c  =  816,  «  =  100°41'. 

d.  Given  a  =  635,  c  =  341,  0=67°  38'. 

e.  Given  c  =  184,  6  =  295,  «=130°54'. 

6.  Discussion  of  checks  and  methods.  —  The  procedure  by 
logarithms  as  with  physical  measurements  involves  numerous 
approximations.  That  two  values  of  log  c,  in  the  check  and  in 
the  solution,  or  by  two  different  methods  of  solution,  do  not 
agree  precisely  is  a  frequent  result  of  correct  computation. 
However,  the  disagreement  will  be  within  certain  well-defined 
limits,  depending  entirely  upon  the  range  (number  of  places) 
of  the  logarithm  tables  which  are  used  both  for  numbers  and 
for  angles ;  the  tabular  difference  should  be  noted,  mentally, 
and  any  discrepancy  between  check  and  computation  should 
be  examined  as  to  its  effect  upon  the  value  of  the  computed 
quantity.  Thus  no  error  in  our  computation  (page  273)  accounts 
for  the  difference  between  log  c  =  2.2068,  and  log  c  =  2.2067, 
nor  does  this  here  affect  the  value  of  c.  However  the 
angle  of  86°  13'  is  so  near  to  90°  that  the  tangent  grows  very 
rapidly ;  the  tabular  difference  here  is  large,  and  might  easily 
affect  our  result,  through  the  inevitable  inaccuracy  of  ordinary 
interpolation  in  this  neighborhood. 


SOLUTION  OF  TRIANGLES  275 

7.  To  determine  the  angles  of  a  triangle  when  the  three  sides 
are  given  :  abc  type.  — 


tan    a  -     r 

V(8  _  a)(s  _  6)(s  —  c) 

2         s-  a 
tan  *  8  —     r 

s 
«4-&  +  c 

2        s-b 
tan    -v  —     r 

2 

2   '         8-C 

a  = 
b  = 

Check.            a  = 
7  = 

28  = 

8  = 

s  —  a  = 

a  +  /3  4-7  = 
log  (s  -  a)  = 
+  log  (s  -  6)  = 
+  log  (s  -  c)  = 

-  log  s  = 

Check  by  noting  sum  of  s  —  a, 
s  —  6,  s  —  c  which  equals  s. 
logr  = 
-  log  (s  -  a)  = 

log  r2  = 
logr  = 

logr  = 
-log  («-&)  = 

log  tan  J  a  = 

log  tan  \  /3  = 

«=  0  = 

logr=  logr  = 

—  log  (s  -  c)  = log*  = 

log  tan  i  7  =  1°&  -A-  — 

7  = 

Complete  the  solution  of  a  problem  of  this  type,  form  as 
above,  taking  a  =  4320,  6  =  6840,  and  c  =  8630. 

TIMING   EXERCISES 

1.  Employing  the  form  of  solution  as  above,  solve  the  follow- 
ing four  problems  for  the  angles  a,  ft,  and  y,  timing  yourself ; 
write  down  the  complete  form  necessary  for  the  solution  of 
each  problem  before  using  the  logarithm  table. 

a.   Given  a  =  320,       6  =  640,       c  =  580. 
6.    Given  a  =  44.8,       6  =  76.2,       e  =  70.4. 

c.  Given  a  =  4.49,       6  =  8.87,       c  =  9.13. 

d.  Given  a  =  .0624,     6  =  .0688,    c  =  .0731. 


276 


UNIFIED  MATHEMATICS 


2.    Solve  the  following  four  problems  for  «,  /3,  and  y,  taking 
note  of  the  time  required. 

b  =  640.6,      c  =  580.4. 
b  =  7461,      c  =  5395. 
6  =  2.346,      c  =  3.045. 
=  2.34xlO-6,  c  =  2.87xlO-6. 


a.  Given  a  =  320.4, 

&.  Given  a  =  3482, 

c.  Given  a  =  1.835, 

d.  Given  a  =  1.43  x  10"6, 


3.    How  would  the  solution  of  problem  2  d  be  changed  if  a, 
6,  and  c  were  given  as  1.43,  2.34,  and  2.87,  respectively  ? 


PROBLEMS 

Type  problems.     Solve  for  the  other  parts 
required  for  the  solution  of  each  problem. 

a  b  c  a. 

33°  15' 


make  a  note  of  the  time 


1. 

8294 

6788 

2. 

8294 

6788 

3. 

8294 

6788 

4. 

8294 

6788 

5. 

8206 

6. 

8206 

7. 

8206 

6009 

8. 

356 

9. 

.03267 

33°  15' 


9645 


33°  15' 


67C 
33° 


25' 
15' 


33°  15' 


67°  25' 


133° 
64° 


15' 
10' 


235 

.05431  63°  40' 

10.  83  x  106,  67  x  106,  54  x  106. 

11.  Given  two  forces  of  magnitude  384  and  276  acting  at  an 
angle  of  38°  with  each  other.     Find  the  angle  which  the  re- 
sultant  makes  with  the  larger   force  and  the  magnitude  of 

the  resultant.  Note  that  the 
problem  is  simplified  by  re- 
garding the  line  of  the  larger 
force  as  an  axis  of  reference  ; 
the  second  force  adds  to  this 
a  component  276  cos  38° ;  the 
276  sin  38° 


Resultant  of  two  forces 
vertical  component  is  276  sin  38 


tan  <J>  = 


384 +  276  cos  38°' 
the  magnitude  of  the  resultant,  r,  is  by  the  sine  law, 


SOLUTION  OF  TRIANGLES  277 

r  -=  276 

sin  (180° -38°)      sin</>' 

276  sin  38° 

whence  r  = — : 

sin  <£ 

or  r  =  V(276  sin  38°)2  +(384  +  276  cos  38°)2, 

as  hypotenuse  of  a  right  triangle  ;  or 


r  =  V(276)2  +  (384)'  -  2  x  384  x  276  x  cos  38°, 
by  the  cosine  law. 

12.  Given  two  forces  of  magnitude  684  and  450,  acting  at  an 
angle  of  64°.     Find  the  resultant  in  magnitude  and  direction, 
graphically  and  trigonometrically. 

13.  From  an  aeroplane  moving  horizontally  at  rate  of  120 
miles  per  hour  (176  feet  per  second)  a  bullet  is  shot  at  right 
angles  to  the  path  of  the  aeroplane  with  a  velocity  of  2800 
feet  per  second.     What  is  the  resultant  velocity  in  magnitude 
and  direction  ? 

14.  A  cylindrical  trough  of  horizontal  length  20  feet  and 
with  the  ends  semicircles  of  radius  2  feet  each,  contains  how 
many  gallons  of  water  for  1  foot  in  depth,  for  1^  feet,  for  2 
feet? 

15.  A  typical  oil  tank  is  30  feet  long  and  has  a  diameter  of 
8  feet.     Compute  the  volume  in  barrels  (see  page  94)  for  each 
foot  of  depth.     Do  not  carry  beyond  tenths  of  a  barrel. 

16.  What  angle  does  ^  =  2#  +  12  make  with  the  #-axis? 
What   angle  does   3  y  —  4  #  —  20  =  0   make  with  the  a^axis  ? 
Find  the  area  and  the  other  angle  of  this  triangle,  formed  by 
the   two   lines   and   the   avaxis,  by  trigonometrical  methods. 
Find    the    angle    between    the    two    Hues    by    the    formula, 

tan  <f>  =  —  ^  —*-    and   check.     Find   the   area  by    analytical 
1 


methods  to  check  upon  the  trigonometrical  solution. 


278 


UNIFIED  MATHEMATICS 


Moving  aero- 
plane 


17.  In  the  figure  AB  represents  the  move- 
ment  in  30  seconds  of  an  aeroplane  moving 
parallel  to  OX  at  rate  of  120  miles  per  hour ; 
Z  AOX  is  measured  as  83°  30'  and  Z  BOX  is 
measured  as  74°  48' ;    find   the    distance    OB 
and  the  position  of  the  aeroplane  at  the  end 
of  the  next  30  seconds  if  it  continues  on  its 
present  course. 

18.  In  problem  17  discuss  the  percentage 
effect  of  an  error  of  1°  in  angle  A  OX  and  in 
angle  AOB,  respectively. 

19.  Given  that  observers  A  and  B  at  the  ends 
of  a  battleship  340  feet  long  observe  an  object 
0  at  angles  of  106°  30'  and  74°  48'  respectively 
with  the  line  AB.     Find  BO  and  AO.      Solve 
this  problem  also  graphically. 


20.  A  and  B  are  observation  stations  on  the  shore,  10  miles 
apart  and  may  be  assumed  to  be  on  an  east  and  west  line ;  R, 
the  battery,  is  three  miles  east 

of  A  and  one  mile  south.  Given 
that  a  battleship  P  is  observed 
from  A  at  an  angle  of  68°  12' 
and  from  B  at  an  angle  of 
59°  02'  with  AB ;  find  the  coor- 
dinates of  P;  find  the  range 
from  R  and  the  angle  made  by 
RP  with  the  east  and  west  line ; 
find  the  distance  using  the  for- 
mula for  the  distance  between 
two  points.  This  problem  is 
solved  in  actual  practice  graph- 
ically on  large  plotting  boards.  Using  i  inch  to  the  mile,  how 
closely  could  you  approximate  the  distance  ? 

21.  In  the  preceding  problem,  suppose  that  the  observa- 
tions reported  at  the  end  of  1  minute  are  from  A,  -(38°  08',  and 


Shore  battery  observations 


SOLUTION  OF  TRIANGLES 


270 


from  B,  59°  12',   locate   the   direction   of   movement  of   the 
ship. 

22.    If  the   battleship   of  the   two  preceding   problems   is 
600  feet  long  and  broadside  to  A,  what  angle  does  it  subtend 


23.  When  guns  are  tested  at  Sandy  Hook  or  other  proving 
grounds,  the  actual  range  for  any  angle  of  elevation  of  the 
gun  is  obtained  by  coincident  observation  of  the  splash  of 
the  shell  by  several  observers  located  in  towers  along  a  line 
roughly  parallel  to  the  line  of  fire  of  the  projectile.  The 


li'ii 


e  ^  Iffiire- 


-E- 


Observation  towers  at  proving  grounds 

shell  is  loaded  with  a  slight  charge  of  high  explosive  in  order 
to  give  a  splash  of  some  magnitude.  For  convenience  in  our 
figure,  we  have  assumed  the  towers  on  a  north  and  south  line 
spaced  as  indicated ;  the  height  of  the  tower  is  regarded  as 
negligible.  Given  that  observers  at  A,  B,  C,  D,  and  E  observe 
the  angle  of  the  splash  with  the  north  and  south  line,  the 
azimuth  angle,  as  2°  20',  21°  24',  27°  16',  41°  35',  and  68°  54', 
find  the  distance  and  direction  of  S,  the  splash,  from  G,  the 
gun,  which  is  400  yards  due  east  of  A.  Note  that  any  two 
observers  give  the  position  of  S ;  the  substantial  agreement 
of  three  observers  is  taken  as  sufficient.  Compare  the  solu- 
tions. Determine  the  coordinates  of  S  with  respect  to  a 
horizontal  axis  through  BCDE  and  a  vertical  axis  through 
AG.  Assuming  that  the  gun  was  pointed  south  the  deflection, 


280  UNIFIED  MATHEMATICS 

to  the  east  here,  is  termed  the  drift ;  it  is  due  to  wind  and 
other  causes  ;  determine  this  angle. 

It  is  not  essential  that  the  towers  be  in  a  straight  line ;  the 
distance  and  direction  of  lines  between  adjacent  towers  is 
carefully  measured.  A  large  plotting  board  is  lised  to  obtain 
a  graphical  solution. 


London  Bridge 

Five  pure  elliptical  arches  ;  central  one  152  feet  by  37  feet  10  inches, 
and  the  others  140  feet  by  37  feet  2  inches. 


CHAPTER   XVIII 

THE   ELLIPSE 

1.   Parametric   equations    of    an    ellipse.  —  The    parametric 
equations  of  the  circle,  with  center  at  the  origin 

x  =  r  cos  6, 

y  =  r  sin  0, 

and  of  the  circle  with  center  at  (h,  Jc) 
x  —  li  =  r  cos  0, 
y  —  Jc  =  r  sin  0, 


if  modified  to  read, 


and 


x  =  T!  cos  6, 
y  =  r.2  sin  0, 

—  h  =  ri  cos  0, 

—  k  =  r  sin  6 


respectively,  give  a  curve  which  is  closely  related  to  the  circle. 
This  curve  is  called  the  ellipse ;  rt  and  r2  are  called  the  major 
and  minor  semi-axes  of  the  ellipse,  and  the  circles  obtained 

281 


282 


UNIFIED  MATHEMATICS 


with  radii  i\  and  r2  are  called  the  major  and  minor  auxiliary 
circles  of  the  ellipse. 


An  ellipse  with  its  major  and  minor  auxiliary  circles 

x  —  TI  cos  6    .  x  =  r2  cos  0    . 

.       gives  the  larger  circle.  gives  the  smaller  circle. 

yt  =  TI  sin  0  &  y  -  r-i  sin  0  5 

x  =  TI  cos  e  . 

„  gives  the  ellipse. 
y  —  r2  sin  0  ° 

P,  Q,  and  R  are  called  corresponding  points. 

Eliminating  0  between  the  two  parametric  equations  of  the 
ellipse  gives 

—  +  £-  =  cos2  0  +  sin2  0  =  1. 

fiz  T  r22 

Writing,  as  is  customary,  a  and  6  for  rt  and  r.2  this  equation 
becomes 


THE  ELLIPSE  283 

If  the  ellipse  had  the  center  (fi,  k)  and  the  two  values  a  and  6 
corresponding  to  rt  and  r.2,  the  equation  of  the  ellipse  would 
be  written 

(x  —  h)  =  a  cos  0, 

(y  —  Jc)  =  b  sin  6, 
in  parametric  form  ;  and 


a2  62 

in  so-called  standard  form.  If  we  assume,  as  is  sometimes 
convenient,  that  a  is  the  radius  of  the  larger  circle,  we  would 
have 

x  —  h  =  b  cos  6, 

y  —  k  =  a  sin  0, 


&2  a2 

as  the  equation  of  an  ellipse  whose  major  axis  is  vertical. 
Note  particularly  that  the  terminal  side  of  the  angle  B  does 
not  pass,  in  general,  through  the  point  on  the  ellipse  which 
corresponds  to  that  angle,  but  the  angle  0  is  made  by  a  line 
passing  through  the  corresponding  points  on  the  major  and 
minor  auxiliary  circles. 

PROBLEMS 

1.  Construct  the  ellipse 

x  =  10  cos  0, 

y  =  6  sin  6, 

by  finding  the  points  on  the  ellipse  given  by  0  =  0°,  30°,  45°, 
60°,  90°,  and  the  corresponding  points  in  the  other  quadrants. 

2.  Construct  the  ellipse 

x  =  10  cos  6, 

y  =  6  sin  0, 

using  corresponding  points  on  the  major  and  minor  auxiliary 
circles  to  locate  points  on  the  ellipse. 


284  UNIFIED  MATHEMATICS 

3.  How  does  the  ellipse 

x—  3  =  10  cos  0, 
y  +  2  =  6  sin  0, 
differ  from  the  preceding  ellipse  ? 

4.  Locate  10  points  011  the  ellipse 

x  —  3  =  6  cos  6, 

y  +  2  =  10  sin  6, 
and  draw  the  curve. 

5.  Prove  that  a  is  the  largest  value  and  b  the  smallest  of  a 
line  from  0  to  any  point  on  the  ellipse.     This  gives  the  reason 
for  the  names,  major  and  minor  axes. 

6.  Show  that  every  chord  of   the  ellipse  through  0  is  bi- 
sected ;  O  is  the  center  of  the  curve. 

7.  The  five  arches  of  the  London  Bridge  are  of  true  elliptical 
shape;   the  central  arch  has  the  width  152  feet  (=2 a)  and 
the  height  37  feet  10  inches  (=6).     Find  the  equation  and 
plot  12  points  on  this  arch ;  b  may  be  taken  as  38  feet.     The 
adjacent  arches  are  of  length  140  feet  and  height  37  feet  2 
inches.     Write  the  equations.     In  each  case  take  the   major 
axis  as  avaxis  and  minor  axis  as  y-axis. 

2.    Properties  of  the  ellipse.  —  The  equations 
x  =  a  cos  0, 
y  =  b  sin  9, 

wherein  a  >  &,  can  represent  any  ellipse  whatever,  when  the 
axes  of  the  ellipse  are  taken  as  the  axes  of  reference.  The 
geometrical  peculiarities  of  this  ellipse  will  be  characteristic 
of  all  ellipses,  provided,  of  course,  that  no  limitation  is  placed 
upon  a  and  b  (except  that  a  may  be  taken  as  greater  than  b). 
Each  ordinate  of  the  ellipse 

x  =  a  cos  0, 
y  =  b  sin  0, 


THE  ELLIPSE 


285 


is  a  constant  proportional  part  of  the  corresponding  ordinate 
in  the  major  auxiliary  circle, 

x  =  a  cos  0, 
y  =  a  sin  6. 


An  ellipse  with  its  major  and  minor  auxiliary  circles 

x  =  TI  cos  9    .  x  =  T-L  cos  6    . 

.       gives  the  larger  circle.  „  gives  the  smaller  circle. 

y  =  TI  sin  6  °  y  =  rz  sm  0  ° 


i  .        ,,       ... 

gives  the  ellipse. 
y  =  rz  sm  6  & 

P,  Q,  and  E  are  called  corresponding  points. 

Take     any    point     (a  cos  0,  b  sin  6)     on     the     ellipse,     the 
corresponding    ordinate    in    the    circle   is    ye  =  a  sin  0  ;    but 

yt=b  sin  0  =  -  (a  sin  0)  =  -  •  yc.     Conversely,  if  the  ordinate  of 

a  a 

any  point  on  a  given  curve  is  always  a  constant  proportional 


286 


UNIFIED  MATHEMATICS 


part  of  the  ordinate  of  a  corresponding  point  on  a  given 
circle,  for  equal  values  of  the  abscissa,  the  curve  is  an  ellipse ; 
the  ratio  need  not  be  less  than  unity  as  the  figure  clearly 
indicates,  a  vertical  ellipse  being  represented  when  the  ratio 
is  greater  than  unity.  This  property  of  this  ellipse  is  equally 
true  of  any  ellipse,  replacing  the  terms  ordinate  and  abscissa, 

where  given  or  implied, 
by  perpendiculars  to  the 
major  and  minor  axes,  re- 
spectively, of  the  curve. 

If  the  major  auxiliary 
circle  is  rotated  about 
OX  as  an  axis  through 

£+      an  angle  a,  cos  a  =  ~ ,  the 
a 

projection  upon  the  plane 
of  the  original  position 
will  be  the  ellipse 

x  =  a  cos  6, 
y  =  b  sin.0; 

evidently  the  x  of  any 
point  on  the  projected 
curve  may  be  taken  as  the 
x  of  a  point  on  the  project- 
ing circle,  or  x  =  a  cos  6. 
The  ordinate  on  the  pro- 
jected curve  is  in  a  con- 
stant ratio,  cos  «  =  - ,  to 

a 
Elliptical  section  of  a  circular  cylinder          t^e  ordinate  on  the  circle. 

Further  any  plane  section  of  a  cylinder  with  circular  base  is 
an  ellipse,  for  with  the  same  abscissas,  the  ordinates  of  the 

curve  of  section  bear  a  constant  ratio  -    —  to  the  correspond- 


cos  a 


ing  ordinates  on  the  circular  base. 


THE  ELLIPSE  287 

Let  the  plane  cut  the  axis  of  the  cylinder  in  0,  and  let 
BOX,  on  the  diagram,  be  the  intersection  of  the  cutting 
plane  with  the  plane  of  the  circular  section  through  0  (paral- 
lel to  the  base  of  the  cylinder).  Note  that  angle  PMQ  is 
constant. 

Similarly  if  a  circular  cone  is  cut  by  a  plane  cutting  all  the 
elements  of  the  cone,  the  section  formed  is  an  ellipse;  the 
geometrical  proof  is  rather  too  complicated  to  give  here  but  an 
analytical  proof  is  indicated  in  Chapter  XXXII,  Section  6. 
The  ancient  Greeks  studied  the  properties  of  the  ellipse  entirely 
from  the  point  of  view  of  the  curve  as  a  plane  section  of  a 
cone.  The  Greek  theoretical  work  concerning  the  properties 
of  conic  sections  made  it  possible  for  Kepler  to  discover  that 
the  path  of  the  earth  about  the  sun  is  an  ellipse,  and  for 
Newton  to  formulate  the  law  of  gravitation. 

The  properties  mentioned  are  intimately  connected  with  the 
applications  of  the  ellipse  in  engineering  problems. 

From  the  property  of  the  ordinates  it  follows  that  the  area 

of  an  ellipse  is  -  •  Ti-o2  =  nab.     A  =  irab. 
a 

The  proof  is  strictly  by  a  "  limit  process,"  and  may  be  made 
reasonably  evident  by  dividing  the  semi-axis  into  25,  50,  100,  — 
equal  parts  and  drawing  two  series  of  rectangles  on  these 
equal  subdivisions  about  the  corresponding  ordinates  to  fall 
entirely  within  (or  entirely  without)  the  ellipse  and  the  circle, 
respectively.  As  the  subdivisions  are  increased  in  number  the 
one  series  of  rectangles  has  the  area  of  the  quarter-ellipse  as  a 
limit ;  it  differs  from  this  area  never  by  an  area  as  great  as  the 
rectangle  of  height  b  and  base  one  subdivision ;  so  also  the 
sum  of  the  second  series  of  rectangles  has  the  quarter-circle 
as  a  limit,  never  differing  from  it  by  an  area  as  great  as  the 
rectangle  of  height  a  and  base  one  subdivision ;  but  the  sum 

of  the  series  of  smaller  rectangles  always  equals  -  times  the 

a 

sum  of  the  series  of  larger  rectangles  since  the  bases  are  equal 


288 


UNIFIED  MATHEMATICS 


and  the  altitude  of  any  one  of  the  smaller  is  -  times  the  alti- 

a 

tude  of  the  corresponding  one  of  the  larger ;  evidently,  then, 
the  limits  of  these  sums  are  in  the  same  ratio.  In  the  dia- 
gram above  the  ruling  of  the  paper  divides  the  semi-major 
axis  into  25  parts,  and  gives  the  rectangles. 


The  Colosseum  in  Rome 

A  famous  elliptical  structure,  615  feet  by  510  feet,  by  159  feet  high  ; 
the  arena  is  an  ellipse  281  by  177  feet. 

A  rectangle  of  dimensions  2  a  and  2  b  may  be  circumscribed 
about  the  ellipse.  The  sides  of  this  rectangle  are  tangents  at 
the  vertices  of  the  ellipse ;  the  middle  lines  parallel  to  the 
sides  are  the  axes  of  the  ellipse ;  the  horizontal  sides  of  this 
rectangle  cut  the  major  auxiliary  circle  in  points  which  cor- 
respond to  four  symmetrically  located  points  on  the  ellipse ; 


sin  0  =  ±  - ,  cos  0  =  ±  \/l  —  —  =  ±  ^— ,  whence 

a  *         a2  a 


x=  ±  Va2—  b2,  y  =  ±  — ,  are  the  coordinates  of  these  points  on 
a  

this  ellipse.     The  points  on  the  major  axes  ( ±  Va2  —  ft2,  0) 
are  the  foci  of  the  ellipse,  and  enjoy  special  properties  with 


289 


respect  to  points  on  the  ellipse, 
the  path  of  the  earth. 


The  sun  is  at  one  focus  of 


A  O  A< 

PF  =  e  •  PZ  defines  the  ellipse 
The  constant  e  is  called  the  eccentricity. 


3.  Standard  definition  of  the  ellipse.  —  The  locus  of  a  point 
which  moves  so  that  its  distance  from  a  fixed  point  called  the 
focus  is  in  a  constant  ratio,  e,  less  than  1,  to  its  distance  from 
a  fixed  line  called  the 
directrix,  is  an  ellipse. 

Let  the  fixed  point  be 
Fand  D'D  the  fixed  line. 
Through  F  drop  a  per- 
pendicular to  the  direc- 
trix D'D  meeting  it  in 
R  and  use  this  line  as 
cc-axis.  By  definition, 
then,  the  ellipse  will  be  the  locus  of  a  point  which  moves  so 
that  PF  =  e  •  PZ,  wherein  PZ  is  the  perpendicular  from  P  to 
the  directrix.  Two  points  of  our  curve  will  be  found  to  lie 
upon  the  avaxis  ;  the  two  points  are  the  points  A  and  A'  which 
divide  the  segment  FR  internally  and  externally  in  the  ratio  e 
(taken  as-|  on  our  figure).  Take  the  middle  point  of  A'  A  as 
the  origin,  designating  A'  A,  which  is  a  fixed  length  dependent 
upon  FR  and  e,  by  2  a.  Then  OA  =  OA'  =  a. 

AF=e>AR.  • 
A'F=e-A'R. 

AF+  A'F  =  e(AR  +  A'R). 
2  a  =  e(AR  +  OA'  + 


=  2  e  -  OR,  whence 


A'F-  AF=  e(A'R  -  AR). 
20F=2ae. 
OF=ae. 

F  is  (ae,  0)  and  D'D  is  x  —  -  =  0  ; 

e 


290  UNIFIED  MATHEMATICS 

PF=  e  •  PZ,  taking  P  as  (x,  y),  gives  in  analytical  language,  or 
formulas, 

V(«  -  ae)2  +  (y  -  O)2  =  e  .  (x  -  -  }  =  (ex  -  a). 


x2  -f  «2e2  +  i/2  =  e2£2  +  a2. 


/}-  ,; 

-H  --  ^  -  =  1. 
a2     a2(l-e2) 

Let  ft2  =  a2(l  -  e2), 


— 
a2 


whence  —  +  ^  =  1.  (1) 

2       2 


But  this  equation  is  satisfied  by  every  point 

x  —  a  cos  0, 
y  =  b  sin  0, 

determined  as  we  have  indicated  above,  and  conversely  ;  our 
two  definitions  are  equivalent  to  each  other. 

The  form  of  equation  (1)  shows  that  the  curve  is  symmetrical 
with  respect  to  the  two  lines  which  we  have  chosen  as  axes  ; 
i.e.  since  x  and  —  x  give  precisely  the  same  equation  to  deter- 
mine y,  the  curve  is  symmetrical  with  respect  to  the  y-axis  ; 
similarly  since  y  and  —  y  give  precisely  the  same  values  for  x, 
the  curve  is  symmetrical  with  respect  to  the  cc-axis.  The 
coordinate  axes  are  axes  of  symmetry  of  the  curve. 

By  symmetry  with  respect  to  the  y-axis  we  mean  that  if  the 
right  half  of  the  curve  is  folded  over  on  the  y-axis  as  an  axis, 
i.e.  revolved  about  it  as  an  axis  (or  axle)  through  an  angle 
of  180°,  the  two  sides  will  coincide  throughout.  Hence 
corresponding  to  Fl  and  D'D,  there  is  another  focus  F2  and  a 
corresponding  directrix  DZD'2,  enjoying  precisely  the  same 
properties  with  regard  to  the  curve  as  F:  and  D'D. 


THE  ELLIPSE 


291 


By  symmetry, 
and  since 

Similarly, 
and,  since 


^  =  e  •  P&,  P2F2  =  e  •  P2Z». 


=  e 


tag 


Kir 


Symmetry  of  the  ellipse  with  respect  to  its  axes 

4.  Sum  of  the  focal  distances  constant.  —  Taking  any  point 
PI(XI,  yi~)  on  the  ellipse,  the  focal  distances  Pi-F\  and  P\F2  are 
equal  to  a  —  exv  and  a  +  ex1}  respectively. 

For  PiFi  =  e  •  PtZ  =  eixl J  =  ex±  —  a,  which  is  a  negative 

V          ej 
distance  since  PI  and  0  are  upon  the  same  side  of  the  line 


=  -  (exv  +  a).     As 


D'D.     Similarly  P&  =  e  -  P&  =  e 


positive  values  P^  and  PjF2  are  a  —  exi,  and  a  +  exl  ;  these 
may  also  be  derived  by  the  formula  for  the  distance  between 
two  points,  noting  that  (xl}  y^)  is  on  the  curve  ; 


PF:  -f  PF2  =  a  —  exl  +  a  -\-  exi  =  2  a. 


292  UNIFIED  MATHEMATICS 

5.   Right  focal  chord.  —  When  x  =  ae, 


?/2  =  52(1  _  e2)  =  -,  since  &2  =  a2(l  -  e2), 

jji  2  52 

y  =  ±—,  giving as  right  focal  chord. 

a  a 

7  2 

The  value  y  =  —  is  the  value  of  y  obtained  in  the  parametric 
ct 

T  Tft 

form  when  sin  6  =  - ,  ?/  =  6  sin  0  =  —  • 

a    '  a 


ft2  / • 

x  =  a  cos  6  =  ±  a-%/1  — — «s  ±  Va2  —  62  =  ±  ae, 

showing  that  the  focus  as  we  have  now  defined  it  coincides 
with  the  focus  as  first  defined. 

6.    Standard  forms  of  the  ellipse.  —  The  equation  — (-  ^-  =  1 
may  be  interpreted  geometrically, 


=  1, 


in  which  CM  and  MP  represent  the  distances  cut  off  from  the 
center  on  the  major  and  minor  axes  of  the  ellipse  by  the  per- 


THE  ELLIPSE  293 

pendiculars  to  the  axes  from  any  point  on  the  ellipse ;  CA  and 
CB  are  the  lengths  of  the  semi-major  and  semi-minor  axes. 

Given  a  horizontal  ellipse  with  center  at  (h,  k)  and  axes  a 
and  b,  the  relation 

CM*     "Ml*  _  1 

CA2      CB2~ 

becomes 


a2  b2 

=  x  —  h  and  MP  —  y  —  k. 

Similarly  a  vertical  ellipse  with  center  (h,  k)  and  axes  a  and 
&,  respectively,  has  the  equation 

(y  —  k)2  ,  (x  —  h}2 

— «-  +  i '—  =  1.  since 

a2  b2 


CM  is  here  y  —  k  and  J£P  is  x  —  h. 

m        f  (x-hY  ,  (y—k)2  (y  —  k)2  .  (x-hy      1 

Type  forms  :  ^  -  ^-  +  Vz  --  <L  =  \  •   \M.  --  L  +  ^  --  L  =  l. 

a2  62  a2  62. 

&2  =  a2(l  -  e2)  ; 

.         [horizontal    ..  ..       ly  —  k  =  0 

foci  of    ^        ,  .     .       ellipse  on  the  line  { 

\  vertical  [  x  —  h  =  0 

at  a  distance  ae  =  Va2  —  62,  from  the  center  (h,  k)  ; 

right  focal  chords  extending  a  distance  —  {  vei 

a  |  horizontally 


from  the  foci  on  either  side  of  the  maior  axis       ~    ~~   ' 
7.   Limiting  forms  of  the  ellipse  equations.  — 

ly>   _    7.\2 


_ 

+  ^  -  *-  =  l  represents  an  ellipse. 
a  62 

•  —  t-  +  ^^  —  '—  =  0   represents   a    point   ellipse   or    two 

imaginary  straight  lines  through  (h,  k)  ;  the  only  real  point 
which  satisfies  this  equation  is  the  point  (h,  k). 


294  UNIFIED  MATHEMATICS 

(y.   Jf\2  (ft  Jf\2 

i '*-  +  ^ 1-  =  —  1    represents   an  imaginary    ellipse ; 

only  imaginary  values  of  x  and  y  can  make  the  sum  of  the  two 
squares  on  the  left  equal  to  —  1. 

The  equation  of  any  ellipse  with  axes  parallel  to  the  coordi- 
nate axes  may  be  written, 


=  k,  wherein 
a2  6 

ka2  and  kb*  represent  the  squares  of  the  semi-axes.  As  k 
approaches  zero  the  ellipse  diminishes  in  size,  and  when  k  =  0 
the  equation  represents  the  point  (x1}  y^;  when  k  becomes 
negative  the  ellipse  becomes  imaginary. 

8.    Illustrative  problems. — 

a.   Plot  the  ellipse  4  xz  +  16  x  +  9  y*  —  18  y  —  75  =  0. 

First  write  in  form  to  complete  the  squares, 

4(x2  +  4  x        )  +  9(y2  —  2  y        )  =  75. 
Complete  squares : 

4(x2  +  4x  +  4)+  9(j/2  —  2  y  +  1)  =75  +  16  +  9. 
4(x  +  2)2  +  9(y  -  1)2  =  100. 

Write  in  standard  form  : 

(x  +  2V 
25 


ft2  _  lOQ  _  20 
a  ~  45  ~  9 

Plot  the  center,  extremities  of  major  and  minor  axes  ;  foci ;  extremities 
of  right  focal  chords  ;  at  least  one  further  point,  obtained  from  the 
original  equation  and  selected  so  as  to  give  a  point  approximately  midway 
between  the  extremity  of  a  focal  chord  and  the  corresponding  extremity 
of  the  minor  axis  ;  by  symmetry  three  other  points  are  obtained.  In  this 
case  x  =  0  gives  desirable  further  points. 

9  ±  \/81  +  675 


=  1  ± 

=  1  ±  K27.5) 

=  1  ±  3.06 

=  4.06  or  -  2.06. 


THE  ELLIPSE 


295 


b.  Plot  the  elliptical  arch  of  a  bridge,  arch  125  feet  wide  and 
37  feet  high.  Plot  points  for  every  10  feet  of  the  span ;  and 
compute  to  TX7  of  1  foot. 


|||g|jap     :  ^ 

;  ;  1   ;   hi       & 

-I  _l  -,,(--  

1                   IE    3 

? 

:fe:r                                :I      1  -i:" 

2 

..®Ssg;  :|±: 

2 

"~~.                                                     X.1        5> 

^                                                    gj  —    ^    

/ 

"-                                                     ._ 

...j  

-  >    -•                                            K              S 

»--          i  -- 

-v  ti^-^l  

-      S  i    II    i  — 

-,    .-/    in      jo      :,,,     _4o      :,(.)      60~" 

Elliptical  arch,  125  foot  span  by  37  feet  high 

Scale,  1  inch  to  40  feet.     Horizontal  measurements  are  from  the  middle 
of  a  quarter-inch  square. 


*      _L  y   —  \ 

(62.5)2      372 
=  (62. 5)2;  &2=372;  ae  = 


-372;   ^-  =  J^_. 
a      62.5 


62.52 


Here  compute  372  and 


2/2  =  372  _  /JLV  .X2, 

V62.5>/ 
/_37_y 

\62.5y 


,  and  multiply  the  latter  by  x2  =  100, 


400,  900,  1600,  2500,  and  3600  ;  extract  the  square  root  of  the  difference  ; 
use  four-place  logarithms. 

log  37  =  1.5682 
log  62.5  =  1-7959 
log  quot.  =  9.7723  —  10 
log  quot.2  =  9.5446  —  10 


(— V- 

\Q2.b) 


.3504 


=  1369  -      35.04  ;  yw  =  -f-  36.51 


2/202  =  1369  -    140.  16 


=  +  34.93 


2/ao2  =  1369  -  315.36  ;  y30  =  +  32.46 
2/402  =  1369  -  560.64  ;  y^  =  +  28.43 
I/so2  =  1369  -  876.0  ;  yx  -  +  22.20 
yeo2  =  1369  -  1261.44  ;  y«  =  +  10.37 


296  UNIFIED  MATHEMATICS 

EXERCISE. — Draw  the  major  auxiliary  circle  on  half-inch 
coordinate  paper,  and  compute  corresponding  ordinates  in  the 

37 

preceding  arch  as  -f   -  of  the  ordinates  on  the  circle ;    e.g.  for 
62.5 

S7 

x  =  10,  find  y  graphically  on  the  circle  and  multiply  by  -  —  • 

02.  o 


PROBLEMS 

1.  Plot  the  ellipse  9  x2  +  36  a  +  #2  —  6  ?/  =  0  ;  what  are  the 
coordinates  of  the  foci  ? 

2.  Plot  one  quarter  of  the  ellipse  -  —  +  -^—  =  1. 

1472     592 

3.  Plot  an  elliptical  arch,  width  233  feet,  height  73  feet, 
plotting  at  least  10  points  spaced  at  20-foot  intervals  from  the 
center.     These  are  the  dimensions  of  the  arch  of  the  Walnut 
Lane  Bridge  in  Philadelphia;  the  arch  is  approximately  an 
ellipse. 

4.  Plot  the  upper  half  of  an  ellipse,  giving  a  vertical  ellip- 
tical arch  13  feet  1|  inches  high,  and  9  feet  2£  inches  wide. 
This  represents  the  upper  portion  of  an  elliptical  sewer  used 
in  the  city  of  Philadelphia. 

5.  What  limitation  is  there  upon  the  values  of  A  and  B  if 
the  equation,  A&  +  By2  +  2  Gx  +  2  Fy  +  C  =  0,  is  to  represent 
an  ellipse  ? 

6.  Put  the  following  equations  in  standard  form,  complet- 
ing the  square  first  and  reducing  to  standard  form  by  division. 
Time  yourself. 


y*  —  6y—4:3=  0. 

c.  ox*-17x  +  10y2-  100  =  0. 

d.  5*2  +  127/2-117  =  0. 


7.   Plot  the  preceding  five  ellipses,  choosing  an  appropriate 
scale.     Plot   the   extremities   of   major  and  minor  axes,  the 


THE  ELLIPSE  297 

extremities  of  the  right  focal  chords,  and  one  other  point 
obtained  by  computation,  together  with  the  three  points  sym- 
metrical to  the  computed  point. 

8.  Determine  a2  and  ft2  to  one  decimal  place,  in  the  follow- 
ing three  ellipses  : 

a.  17  x2  +  43  y*  =  397- 

b.  5x2-17x  +  10y*-35y  =  0. 

c.  l(x  -  2)2  +  3(y  -  3)2  =  39. 

9.  In  the  three  ellipses  immediately  preceding  determine 

ae,  —  ,  and  -  to  one  decimal  place. 
a  e 

10.  In  each  ellipse  of  problem  8  determine  x  when  y  =  2. 

11.  Using  the  data  of  problems  8,  9,  and  10,  plot  the  three 
ellipses  of  problem  8. 

12.  In  the  ellipse  -^-  +  f-  =  1,  find  the  foci.     What  is  the 

100     36 

distance  of  the  point  whose  abscissa  is  +  3  from  each  focus  ? 
of  the  point  whose  abscissa  is  4,  5,  6,  7,  x±  ? 

13.  Put  the  following  equations  in  standard  form  and  dis- 
cuss the  curves  : 

a.  x*-6x  +  y2+8y-10=Q. 

b.  z2-  6z  +  4?/2+  8    +  11  =0. 

c.  B2- 

d.  x2- 

e.  x2  — 


14.  The  path  of  the  earth  about  the  sun  is  an  ellipse  with 
eccentricity  .01677  ;  this  may  be  taken  as  ^  in  the  following 
computations.  If  the  major  axis  of  the  earth's  orbit  is  185.8 
million  miles,  determine  the  focal  distance,  i.e.  from  the  sun  to 
the  center  of  the  path  ;  determine  also  the  minor  axis.  If  a 
scale  of  one-half  inch  to  fifteen  million  miles  is  taken,  at  what 
distance  will  the  point  representing  the  sun  be  from  the  center 


298 


UNIFIED   MATHEMATICS 


of  the  path  ?     What  will  be  the  difference  in  length  between 
major  and  minor  axes? 


The  path  of  the  earth  about  the  sun 

The  distance  of  the  sun  from  the  center  of  the  ellipse  is  represented 
by  ^5  of  an  inch,  on  this  diagram. 

9.  Tangent  to  ellipse  of  given  slope.  —  (See  page  225.)  To 
find  the  tangent  of  slope  2,  y  =  2  x  +  k  is  solved  as  simultaneous 
with  the  ellipse  equation.  An  equation  whose  roots  are  the 
abscissas  of  the  two  points  of  intersection  is  found  and  the  con- 
dition is  used  that  the  two  points  of  intersection  be  coincident. 

a;2 


-  =  1, 
a*      b2 

y  =  2  x  4-  k. 

bW  +  a?  (4  x*  +  4  kx  +  fc2)  —  a? 
xz  (b*  +  4  a2)  +  4  azkx  -f-  a,2A;2  - 


=  0. 
=  0. 


_ 


±  V4  a4fc2  -  (fr2  +  4  a 


&2  +  4  a2 


-  2 


-  2  «2fc  ± 


4  a2  -  ft2) 


62  +  4  a2 


THE  ELLIPSE  299 


Put  52  +  4  a2  _  fc2  _  o. 

ft2  =  6s  +  4  a2, 


k  =  ±  V62  -f  4  a2, 


y  =  2  a;  ±  V&2  +  4  a2 
are  the  two  tangents  of  slope  2. 

Similarly  the  tangent  of  slope  m  is  obtained  by  solving 


(2)  y  =  mx 

as  simultaneous,  and   writing  the   condition  for  equal   roots. 
Clearing  (1)  of  fractions, 

&2a2  +  a2?/2  -  a2^2  =  0  ; 
substituting  from  (2), 
62cc2  +  a2  (m2a2  -f-  2  Jcmx  +  A;2)  —  a262 

=  (62  +  a2™2)  x2  +  2  fca2wwj  +  (a2A;2  -  a262)  =  0. 
_  -  ka*m  ±  Vfc2a4r/i2  —  (62  +  a2m2)(a2fc2  -  a262) 

62  +  a2m2 

Putting  the  discriminant  equal  to  zero, 
A;2  =  62  +  a2w2. 


A;  =  ±  V62  4-  a2m2,  whence 
y  =  mx  ± 

are  the  two  tangents  of  slope  m  to 


This  method  of  obtaining  the  tangent  applies  to  any  curve 
given  by  an  equation  of  the  second  degree. 

10.  Focal  properties  of  the  ellipse.  —  The  perpendicular  from 
the  focus  (ae,  0)  on  any  tangent  of  slope  m  meets  it  in  the 
point,  whose  coordinates  are  found  by  solving  the  equations  of 
these  lines  as  simultaneous. 


300 


UNIFIED  MATHEMATICS 


The  equations,       y  =  mx  +  Va2™2  +  62,  of  the  tangent, 


and 

may  be  written 


y  —  0  = (x  —  ae),  of  the  perpendicular, 

m 


y  —  mx-  =  V«2m2  +  62. 
my  +  x  =  ae. 


The  perpendicular  from  the  focus  upon  any  tangent  to  an  ellipse  meets 
it  on  the  major  auxiliary  circle 

The  point  of  intersection  satisfies  both  these  equations  ; 
further  it  satisfies  the  equation  obtained  by  squaring  and  add- 
ing both  members  of  each  of  these  equations  : 


my  +      +  ma  =  ara 

=  a2m2  +  62  +  a2  —  62 
=  a2(l  +  m2). 
a2  +  f  =  a2  ; 

hence  the  point  of  intersection  of  the  perpendicular  from  the 
focus  on  any  tangent  lies  on  the  major  auxiliary  circle. 

Note  that  the  above  demonstration  applies  equally  well  to 
the  perpendicular  from  the  other  focus  (—ae,  0)  and  equally 
well  to  the  other  tangent  of  slope  m,  y'=  mx  —  Va2m2  -f-  62. 


THE  ELLIPSE 


301 


11.   Tangent  to  an  ellipse  at  a  point  PI(X{,  y^)  on  the  ellipse.  — 

The  method  outlined  is  general,  being  applicable  to  any  alge- 
braic curve.  The  point  PI(#I,  yi)  on  the  curve,  considered  as 
fixed  during  the  discussion,  is 
joined  to  a  neighboring  point 
PZ(XI  4-  h,  yt  +  ft)  on  the  curve, 
which  second  point  is  then  made 
to  approach  (xl}  y^)  along  the  cwve. 
The  slope  of  the  chord  joining  Px 


-  is  the  slope  of  the  chord 
h 


Jc 

to  P2,  -,  is  found  not  to  change 
h 

indefinitely,  but  is  found  to  ap- 
proach a  definite  limiting  value  as 
h  and  k  approach  zero,  i.e.  as  P2 
approaches  Px  along  the  curve. 
This  limiting  position  of  the  chord 
is  the  tangent  at  Pl ;  this  line  can  be  shown  in  the  case  of 
the  ellipse  (or  any  curve  given  by  an  equation  of  the  second 
degree)  to  cut  the  curve  in  two  coincident  points  at  (x1}  y^) 
and  in  no  other  point. 

The  method  is  applied  in  parallel  columns 

to  a  general  problem  and  to  a  particular  problem. 

Tangent  to  the  ellipse, 


- 

a2     &2 

at  PI(XI} 


. 

25     4  " 

at  Pi(3,  f  ) 


on  the  ellipse. 
Take  the  second  point 

PZ(XI  +  h,  yl  -f-  Jc)  on  curve  P2(3  +  h,  f  +  k)  on  curve 

Substituting, 

+  hy  +  a?(yi  +  A;)'  -  a2&2  =  0. 

4(3  +  /i)2  +  25(f  +  fc)*  -  100  =  0. 

+  2  62*0?!  +  62/i2  +  a2y2  +  2  d^kyl  +  aW  -  a262  =  0. 
But  &V  +  aVi2  -  a262  =  0. 

36  +  24  h  +  4  fc2  +  64  +  80  k  +  25  fc2  -  100  =  0. 


302 


UNIFIED  MATHEMATICS 


i  +  52^2 


Subtracting  (and  canceling)  . 

•,  +  a2&2  =  0.    24  ft44  fc2+80  k  4-  25  fc2=0. 
&(804-25  k}=  -ft(2444  h). 

k=       24  4 4 h 

h         2a?i/i  +  a?k  h         80  +  25fc' 

The  chord  P±PZ  is  given  by : 

y -»i  =  £(*-«!).  y-|  =  7(*-3). 

ft  Q        II 

Since  P2  is  on  the  curve,  the  chord  equation  may  be  written : 


11—11,= 


!  +  Wh 

-!  - 


+ 


24  +  4  h 


,        ox 
(x  —  3). 

v 


Let  P2  approach  PY  along  the  curve  ;  h  and  A;  both  approach 
0,  i.e.  can  be  made  just  as  small  as  you  please.     Thus  if  h  is 

made  .01  in  our  numerical  prob- 
lem, k  will  be  about  —  .003,  which 
can  be  obtained  by  solving  the 
quadratic 


JEfM  -i-  -/r4, 

^jffl^Fi^ 


^ 


for  fc.  It  is  evident  that  the  con- 
stant here  may  be  regarded  as 
24  h  +  7i2  and  that  as  ft  =  0,  this 
constant  approaches  zero,  and  one 
root  of  k  approaches  zero. 

Xote  that  the  second  value  of 
A;  approaches  —  2  y1}  and  corre- 
sponds to  the  fact  that  the  given  value  of  x,  xv  +  h,  is  the 
abscissa  of  two  points,  on  the  upper  and  lower  parts  of  the 
curve,  respectively. 

Since  h  and  k  both  approach  zero, 


-  is  the  slope  of  the  chord  PiP2 


and 


24  +  4ft    =24 

and 
-  80  4  25  fc  =  -  80 


THE  ELLIPSE  303 

and  the  slope  of  the  chord  approaches  more  and  more  nearly, 
and  as  near  as  you  may  please  to  make  it,  by  taking  h  (and 
thus  A;)  small  enough, 

_24 

~80' 

Hence  the  chord  approaches  a  limiting  position,  given  by 

N  8          24, 


which  may  be  written 

a*yiy  =  aV  +  Vx?          y  -  1.6  =  -  .3(x  -  3) 

=  ft262'  or  y-1.6  =  -  .3(x  -  3). 

a?yiy  -  a*&2  =  0  10  >/  +  3*  -  25  =  0. 

40  ?/  +  12*  -100  =  0. 


By  precisely  this  method,  step  for  step,  the  tangent  to 

Ax*  +  By*  +  2Gx  +  2Fy  +  C  =  0,  is  found  to  be 
Axjx  +  Bihy  +  G(x  +  a;,)  -f-  ^(y  +  2/0  +  C  =  0. 

PROBLEMS 

1.  Find  the  tangents  of  slope  -f  2  and  of  slope  —  3  to  each 
of  the  ellipses  in  problem  6  of  the  preceding  set  of  problems  ; 
the  five  problems,  tangents  of  slope  +  2,  should  take  not  to 
exceed  30  minutes  ;  note  that  after  substituting  2  x  +  k  for  y  it 
is  better  procedure,  surer  and  quicker,  to  combine  terms  by  in- 
spection rather  than  to  expand  each  binomial  before  combining. 
This  means  to  pick  out  the  terms  containing  a;2,  for  example, 
and  write  the  sum  of  these  coefficients  directly. 

2.  Draw  at  least  three  of  the  tangents  of  slope  +  2  in  the 
preceding  exercise,  and  three  of  slope  —  3,  each  to  its  conic 
as  previously  drawn.     Find  the  point  of  tangency  algebraically 
and  graphically. 


304  UNIFIED  MATHEMATICS 

3.  Without   plotting  the  ellipse  itself  plot  12  tangents  of 
slope  0,  1,  2,  3,  4,  5,  6,  and  10,  and  of  slope  —  ^,  —  1,  —  3,  and 
—  6  to  the  ellipse  9xz  +  25y2  =  900 ;   note  that  these  give  a 
fair  outline  of  the  ellipse. 

4.  In  each  of  the  ellipses  of  problem  8,  page  297,  find  the 
tangents   at   the    point  whose   ordinate    is   2,  employing   the 
results  of  problem  10  of  the  same  set,  and  using  the  formula, 
Ax&  +  Byvy  +Q(x  +  xl)  +  F(y  +  yt)  +  G'=  0. 

5.  Derive   by  the  method  outlined   in   section   11   of   this 
chapter  the  tangent  to  the   ellipse   9  xz  +  25  y1  =  900   at   the 
point  (8,  3.6)  which  is  on  the  curve  ;  at  (—  6,  4.8)  on  the  curve. 

6.  In  the  ellipse  9  x2  +  25  y1  =  900,  verify  that  the  perpen- 
dicular from  the  focus  upon  any  tangent  to  the  ellipse  meets 
it  on  the  major  auxiliary  circle.     Note  that  the  converse  is  also 
true.     This  gives  a  method  for   drawing  the  tangent  to  an 
ellipse  from  a  point  outside  the  ellipse ;  explain. 

12.  The  tangent  to  an  ellipse  at  a  point  on  the  ellipse  con- 
structed from  the  tangent  to  the  auxiliary  circle.  —  Let  (x1}  y^ 

be  any  point  on  the  ellipse  ;  the  tangent  is  — ^  -+-  ^^  =  1 ;  the 

a2 
a?-intercept,   xi}   of   this   tangent   is    — ,  obtained   by   solving 

XL 

•?L-  -|_  IM.  =  1  as  simultaneous  with  y  =  0.  Evidently  x,  =  — 
a2  b-  Xi 

depends  only  upon  x\  and  a,  not  involving  b  or  y±.  Hence  this 
value  would  be  unchanged  if  b  were  taken  equal  to  a.  The 
tangent  to  the  major  auxiliary  circle  x2  -f-?/2  =  a2  at  (xl}  yz)  on 
the  circle  is  x^x  +  y?y  =  a2,  and  the  intercept  of  this  tangent 

a2 

on  the  ic-axis  is  also   — .     This  gives  the  following  rule  for 
yl 

drawing  a  tangent  to  an  ellipse  at  any  point  on  the  ellipse  : 

Construct  the  major  auxiliary  circle  to  the  given  ellipse;  find 
the  point  P2  on  the  circle  having  the  abscissa  of  the  given  point; 
at  the  point  P2  construct  the  tangent  to  the  circle,  cutting  the 


THE  ELLIPSE 


305 


x-axis  at  T;  the  line  joining  T  to  Pv  on  the  ellipse  is  tlie  tangent 
to  the  ellipse. 


Tangent  to  an  ellipse  constructed  from  the  tangent  to  the  auxiliary  circle  a 
the  corresponding  point 

Note  that  even  if  6  is  greater  than  a,  this  construction  gives 
the  tangent,  but  the  circle  ce2  +  y2=a2  is  then  the  minor 
auxiliary  circle  of  the  given  ellipse. 

13.    The  tangent  to  an  ellipse  bisects  the  angle  between  the 

focal  radii  to  the  point  of  tangency.  —  Let  ^  +  M  =  1    be   the 

a2        62 

tangent  at  (x1}  yt)  which  intersects  the  a;-axis  at  T  (—,  0\ 
22  ^      ' 


Hence, 


a-ex,, 


306 


UX I  PIED  MATHEMATICS 


.-.  PiT  bisects  the  exterior  angle  of  the  triangle 
since  it  divides  the  opposite  side  into  segments  proportional 
to  the  adjacent  sides.  The  normal  bisects  the  interior  angle 
between  the  two  focal  radii ;  if  the  normal  be  drawn,  each  focal 
radius  makes  the  same  angle  with  it. 


Tangent  to  an  ellipse  constructed  from  the  focal  radii  to  the  point  of 
tangency 

Any  ray  of  light  or  sound  striking  a  reflecting  surface  is 
reflected  in  the  plane  of  the  normal  to  the  surface  and  the 
original  ray  in  such  a  way  as  to  make  the  angle  of  incidence 
(i.e.  between  normal  and  original  ray)  equal  to  the  angle  of 
reflection.  Hence  rays  starting  from  FI  in  our  figure  con- 
verge at  F2.  This  is  the  principle  of  "  whispering  galleries," 
in  which  the  rays  of  sound  starting  from  a  point  FI  converge 
at  another  point  F2,  making  audible  at  F.2  whispers  at  Ft ;  at 
intermediate  points  the  conversation  may  not  be  audible  as 
there  is  no  reenforcement  by  convergence. 


PROBLEMS 


1.  In  the  ellipse  9  x2  +  25  y1  =  900,  give  the  two  graphical 
methods  for  drawing  the  tangent  at  the  point  (6,  4.8)  on  the 
ellipse. 


THE  ELLIPSE  307 

2.  In    the    ellipse    25  x2  +  9  y2  =  900,    give    the    graphical 
methods  for  drawing  a  tangent  at  the  point  (—4.8,  6)  on  the 
ellipse. 

3.  Construct  the  three  ellipses  : 

100  +  36  = 

ioo  +  ioo  =  L 
_*!.  HL  y!_  =  i 

100  144 

Draw  the  tangent  to  each  of  these  ellipses  at  the  point 
whose  abscissa  is  +  6 ;  find  the  equation  of  each  of  these 
tangents  and  prove  that  they  intersect  on  the  avaxis. 

4.  Find  the  equations  of  the  two  focal  radii  to  the  point 

x~       v2 

(6,  4.8)  on  the  ellipse   -  —  +  —  =  1 ;  find  the  bisectors  of  the 
100     36 

angles  between  these  focal  radii ;  find  the  bisector  of  the  angle 
which  does  not  include  the  origin,  and  prove  that  it  .coincides 
with  the  tangent  at  (6,  4.8)  to  the  ellipse. 

5.  If  an  elliptical  arch  is  to  be  in  the  form  of  the  upper 
half  of  an  ellipse,  find  the  equation  and  plot  ten  points,  given 
that  the  width  of  the  arch  is  to  be  100  feet  and  the  height  is  to 
be  40  feet. 

6.  If  the  preceding  arch  is  to  have  the  dimensions  as  given, 
but  is  to  be  constructed  as  the  upper  quarter  of  a  vertical 
ellipse,  find  the  equation  of  the  curve.     Note  that  you  have  a 
point  (50,  40)  which  is  to  satisfy  the  equation  of  the  curve 
which  can  be  written  with  only  the  denominator  of  x2  as  un- 
known.    Compare  this  arch  with  the  preceding  one  as  to  beauty 
of  design. 

7.  Find  the  lengths  of  the  ten  vertical  chords  of  the  arch 
in  problem  5,  dropped  from  the  tangent  at  the  top  of  the  arch 
and  equally  spaced  horizontally. 


308  UNIFIED  MATHEMATICS 

8.  By  the  method  of  article  13,  find  the  tangent  to  the 
curve  given  by  the  equation,  xy=15,  at  the  point  (3,  5)  on  the 
curve. 

9.  By  the  method  of  article  9,  find  the  tangent  of  slope  —  2 
to  the  curve  given  by  the  equation,  xy  =  15. 

10.  Write  the  equations  of  the  three  ellipses  of  problem  8, 
page  297,  in  parametric  form. 

OJ2  y2 

11.  In   the   ellipse,    -— —  +  ^r  =  1,  find  where  lines  from 

100        oO 

(50,  30)  inclined  to  the  horizontal  axis  at  angles  of  15°,  30°, 
45°,  and  60°  respectively,  cut  the  ellipse.  Find  the  lengths  of 
these  lines  from  (50,  30).  See  problem  5. 

12.  If  in  the  ellipse  of  problem  6,  supporting  chords  are 
drawn  diagonally  between  parallel  vertical  chords,  computed  in 
problem  7,  each  from  the  upper  point  of  the  right-hand  chord 
to  the  lower  point  of  the  left-hand  chord  (on  the  right  side  of 
the  ellipse),  compute  the  lengths  of  these  chords. 


From  Tyrrell's  Artistic  Bridge  Design 
Alexander  III  Bridge  in  Paris 
A  parabolic  arch  ;  span,  107.6  m.;  rise,  6.75  in.;  width,  40  m. 


CHAPTER   XIX 
THE   PARABOLA 

1.  Definition.  —  The  ellipse  has  been 
defined  (page  289)  as  the  locus   of  a 
point  which  moves  so  that  its  distance 
from  a  fixed  point,  the  focus,  is  in  a 
constant  ratio  less  than  one  to  its  dis- 
tance from  a  fixed  line,  the  directrix. 
If  this  constant  ratio  is  taken  equal  to 
one,  the  curve  generated  by  the  mov- 
ing point  is  called  a  parabola. 

PF  =  e  •  PZ,  e  <  1,  defines  an  ellipse. 

PF  =  PZ  defines  a  parabola. 

PF  =  e  •  PZ,  e  >  1,  defines  a  hyperbola. 

2.  Equation  of  the  parabola.— 

Through  the  focus  draw  the  perpendic- 
ular FQ  to  the  directrix  ;  take  this  line 
as  avaxis.  Take  FQ,  which  is  constant, 
as  2  a.  On  the  axis  chosen  only  one 

309 


--Z- 


-F- 


F,  the  focus;  ZZ',  the  di- 
rectrii 

A  point  equidistant 
from  Fand  ZZ'  moves  on 
a  parabola. 


310  UNIFIED  MATHEMATICS 

point  is  found  which  is  on  the  given  curve ;  the  mid-point 
0,  dividing  the  segment  QF  in  the  ratio  1  to  1,  is  such  that  its 
distance  from  F,  the  focus,  equals  its  distance  from  the 
directrix.  Through  O  take  a  perpendicular  to  OX  as  the 
y-axis.  Evidently  F  is  the  point  (a,  0),  and  the  directrix  is 
the  line  x  +  a  =  0. 

Take  P(x,  y)  any  point  which  is  on  the  given  curve,  i.e.  any 
point  equally  distant  from  F  and  from  the  line,  giving 

=  PZ. 


PF  =  V(x  —  a)2  +  y2,  distance  between  two  points. 
PZ  =  x  +  a,  distance  from  a  point  to  a  line. 

Note  --ip  gives  the  distance  as  negative ;   but  it  is  not 

necessary  to  take  account  of  the  sign  as  in  the  simplification 
this  expression  is  squared. 
Equating,  PF  =  PZ,  gives 


V(x  —  a)2  -f  yz  =  x  -\-  a. 

x*  —  2  ax  +  a2  +  y2  =  x2  +  2  ax  +  a2. 

y2-  =  4  ax. 

3.  Right  focal  chord.  —  When  x  —  a,  y  =  ±  2  a,  giving  the 
total  length  of  the  right  focal  chord  as  4  a ;  the  coefficient  of 
x  in  y2  =  4  aa;  represents  the  length  of  the  right  focal  chord. 

4.  Geometrical  properties.  —  The  curve  is  symmetrical  with 
respect  to  the  x-axis,  for,  assigning  any  value  to  x,  you  find  for 
y  two  values  ±  V4  ax ;  numerically  equal  but  opposite  in  sign, 
or  lying  symmetrically  placed  with  respect  to  the  OX-line, 
which  consequently  is  here  the  axis  of  the  curve. 

The  y-axis,  x  =  0,  is  tangent  to   this  curve  since,  solving, 

?/2  =  4  ax, 

x  =  0,  as  simultaneous, 

gives  y2  =  0 ;  y  equals  zero  twice,  or  the  two  points  of  inter- 
section  of   x  =  0  with  y2  =  4  ax  are  coincident.      The  point 


THE  PARABOLA  311 

(0,  0)  is  the  point  on  the  axis  of  symmetry,  y  =  0,  which 
corresponds  to  itself ;  this  point  is  called  the  vertex.  The 
line  y  =  0  cuts  the  curve  in  only  one  finite  point,  given  by 
y  =  0,  and  x  =  0  ;  the  other  point  of  intersection  of  y  =  0 
with  the  curve  is  at  an  infinite  distance. 

Given  a  as  positive,  negative  values  of  x  lead  to  imaginary 
values  of  y.  Hence  all  points  on  the  curve  lie  to  the  right  of 
the  tangent  at  the  vertex.  As  x  increases  without  limit,  y 
increases  also  without  limit.  This  curve  extends,  we  may 
say,  to  infinity.  In  plotting  a  parabola,  the  vertex,  the  ex- 
tremities of  the  right  focal  chord,  and  at  least  two  other 
points,  should  be  plotted. 

The  quantity  yz  —  4  ax  is  •  evidently  negative  for  points  in- 
side the  curve,  zero  for  points  on  the  curve,  and  positive  for 
points  outside  the  curve. 

5.  Finite  points  and  the  infinitely  distant  point  on  the  parabola. 
-  To  plot  carefully  yz  =  8  x,  note  that  4  a  =  8,  whence  a  =  2 ; 
indicate  the  vertex  and,  2  units  to  the  right,  the  focus ;  4 
units  (2  a)  above  and  below  the  focus  locate  the  extremities 
of  the  right  focal  chord  which  has  the  length  4  a ;  take  values 
of  x  at  appropriate  distances  from  the  vertex  and  focus  to 
give  the  portion  of  the  parabola  desired  ;  in  yz  =  8  x,  x  =  0,  1, 
2,  3,  4,  6,  and  8  give  sufficient  points  to  plot  the  curve  for  our 
purposes. 

The  parabola  ?/2  =  8  x  is  intersected  by  the  line  y  =  x  at 
(0,  0)  and  at  (8,  8) ;  y  =  ^  x  cuts  this  parabola  at  x  =  0  and  at 
x  =  32  ;  y  =  .1  x  cuts  at  x  =  800,  y  =  80.  These  values  are 
obtained  by  solving  the  equation  y2  =  8  x  as  simultaneous  with 
each  of  the  linear  equations. 

y=.Qlx 
Solvm«>  y>  =  8x 

gives  .0001  xz  =  8  x,  x  =  0  or  x  =  80,000.  If  one  centimeter  is 
taken  as  1  unit,  y  =  .01  x  cuts  the  parabola  y2  =  8  x  at  the 
vertex  and  at  a  distance  of  80,000  cm.,  nearly  ^  mile,  from  the 
vertex.  As  the  line  y  =  mx  moves  nearer  and  nearer  to  the 


312 


UNIFIED  MATHEMATICS 


A  line  parallel  to  the  axis  of  a  parabola  cuts  it  in  a  point  infinitely  distant 

axis  y  =  0,  the  second  point  of  intersection  moves  off  farther 
and  farther,  without  limit.  This  is  the  meaning  of  the  ex- 
pression that  the  axis  of  the  parabola,  and  by  similar  reason- 
ing any  line  parallel  to  the  axis,  "  cuts  the  curve  at  an  infinite 
distance." 

The  methods  given  above  for  plotting  y2  =  8  x  apply  to  any 
parabola  y2  =  4  ax. 

PROBLEMS 
Plot  the  following  parabolas  carefully : 

1.  y2  =  4  x,  from  x  =  0  to  x  =  8. 

2.  yz  =  x,  from  x  =  0  to  x  =  10. 

3.  y-  —  12  x,  from  x  =  0  to  x  =  12. 

4.  y2  =  T\  x,  from  x  =  0  to  x  =  100. 


THE  PARABOLA 


313 


5.  s  =  16 t2,  or  t2  =  TT¥  s,  taking  OS  as  the  horizontal  axis 
and  taking  J  inch  to  represent  10  units  of  s  (distance  in  feet) 
and  |  inch  to  represent  1  unit  of  t  (time  in  seconds).     This 
gives  the  distance  fallen  from  rest  in  time  t  by  a  freely  fall- 
ing body. 

6.  Find   the   intersections   of  y  =  x,  y  =  ^x,   and  y  =  ^x 
with  the  curves  of  problems  1,  2,  and  3. 

7.  Solve  s  =  800 1  with  s  =  16 12 ;  s  =  800  t  is  the  space  cov- 
ered by  a  body  moving  with  uniform  velocity  800  units  per 
second.     What  is  the  physical  meaning  of  the  values  obtained 
for  the  point  of  intersection  ? 

8.  A  mass  rotated  on  a  cord  exerts  a  force  of  tension  on 


the   cord,  F  = 


Given  ra  =  1  pound,  and  r  =  10  feet ; 


draw  the  graph  for  velocities  of  1  to  100  feet  per  second,  taking 
F  on  the  horizontal  axis.  Compute  the  corresponding  number 
of  revolutions  per  minute  for  v  =  10,  20,  50,  and  100  feet  per 
second.  What  is  the  relation  between 
v  and  n,  where  n  is  the  number  of 
revolutions  per  minute  ?  Indicate  on 
the  vertical  scale  a  second  scale  giv- 
ing n. 


6.    Geometrical  interpretation  of 
y2  =  4  ax. 

The  equation  y2  =  4  ax  may  be  inter- 
preted geometrically  as  follows : 

MP-  =  4  a  •  VM, 

the  square  of  the  perpendicular  from 
any  point  on  a  parabola  to  the  axis  is 
equal  to  the  rectangle  formed  by  the 
length  cut  off  on  the  axis  from  the  ver- 


The  square  on  PM  equals 

the  rectangle  with  VM 

and  RFC  as  sides 


tex  by  the  perpendicular,  with  a  constant  line  of  length  4  a, 
the  length  of  the  right  focal  chord. 


314 


UNIFIED  MATHEMATICS 


7.    Standard  and  limiting  forms.  —  Given  that  the  axis  of  a 
parabola  is  parallel  to  one  of  the  coordinate  axes,  the  relation, 


._]____ 
!*:    S:::::::|  i 

leads  to  2 
ard     forn 

MWffllH       ff 

g  I    :      ::::::::::: 

M]?F 

i:||::             ^r? 

parabola. 
Thus  ii 
figure  a 
drawn,  ha 

r   --  i  :Af:-- 

K\  Q  q  VPVT 

!j  ::  J::::|l:i::::::: 

:::fl::t:i^3"S-(t:^ 
::::::::::;:£:::::::::: 

imfii 

axis   para 

E^ctl  rjq/rt  -  -            a-axis     ar 
W1—  :::::::      to  the  ris 
![pp  =  |!E|!      lation 

MP2 

^—                         leads  to 

•    z 

--/-- 

j? 
2 

—  M  ::•:: 

j::::::::::M!^B 
::::::::::::  :|g[g 

:^:::::::::::::        (y  -  5)2  = 
t|r                 Were     F 

ffil::                (*,  *),  Vb 
\\W\~'."=*  ""      os  —  h  sin 
distance  f 

=  4  a  •  VM 


of     the 

e  upper 

abola    is 

F(—  2, 

4  a  =  4, 

to   the 

opening 


A  horizontal  and  a  vertical  parabola 


the    point 
would  be 
it  is  the 
m  a  point 

whose  abscissa  is  h  to 
a  point  whose  ab- 
scissa is  x  ;  similarly  MP  is  y  —  7c  ;  now  when  the  curve 
opens  to  the  right  VM  is  positive  ;  hence  the  equation  is 
(y  —  If}2—  4  a(x  —  h},  with  a  positive.  Were  the  axis  parallel 
to  the  a;-axis,  but  the  curve  opening  to  the  left,  the  equation 
would  be  : 

(y  —  fc)2  =  4  a(x  —  K),  with  a  negative. 

Similarly  the  curve  on  our  figure  which  opens  down  is  given 
by  the  equation  (a?  —  3)2  =  —  S(y  —  2). 

In  the  general  case,  (x  —  7i)2  =  4  a  (y  —  A;)  has  V(h,  fc)  as 
vertex  ;  the  axis  is  x  —  h  =  0  and  is  parallel  to  the  y-axis  ; 


THE  PARABOLA 


315 


the  curve  opens  up  when  a  is  positive  and  down  when  a  is 
negative. 


(y  -  Kf  =  4  a(x  -  h) 


-  A)2  =  4  a  (y  -  K) 


Standard  forms  of  the  parabola  equation 

As  a  approaches  0  in  ?/2  =  4  ax,  the  parabola  approaches 
more  and  more  nearly  to  coincidence  with  the  x-axis  ;  two 
coincident  straight  lines  constitute  a  limiting  form  of  the 
parabola.  As  a  becomes  larger,  the  parabola  approaches  the 
y-axis. 

8.   Tangent  of  slope  m. 

y2  =  4  ax. 
y  =  mx  -\-  Tc. 
+  2  Jcmx  +  fc2  —  4  ax  =  0. 


_  (2  a  —  km)  ±  V(2  a  —  fern)2  — 


ftv- 


whence,  since  A  =  0, 


m 


y  =  mx  +  —  is  the  tangent  of  slope  m  to  y*  =  4  ax ; 
in 


ra2'    m 


is  the  point  of  tangency. 


9.    Tangent  from  an  external  point. — For  any  given  point 
(xi>  y\)  outside  the  curve  two  values  of  m  will  be  found  for 

which    y  =  mx  -\ — :   will  pass  through  (x1}  y^) ;   hence  there 
m 


316 


UNIFIED  MATHEMATICS 


are,  in  general,  two  tangents  which  pass  through  a  given  point 
outside  the  curve.       Ximt  _  yim  +  a  =  0, 


m  = 

For  points  inside  the  curve,  y?  —  4  ax^  is  negative,  and  there 
are  no  tangents. 

10.  Tangent  at  a  point  (xl}  j^)  on  the  parabola.  —  By  the 
method  of  article  9  of  the  preceding  chapter  the  tangent  to 
the  parabola,  yz  _  4  aX}  at  fa}  y^ 

on  the  curve  is  found  to  be 

y^y  =  2  a  (x  -f  «j). 
Similarly  the  tangent  to 

By*  -f  2  Gx  +  2  Fy  +  C  =  0 
is  found  to  be 

By\y  +  G(x  +  ccj)  +  F(y  +  y^)  -\-  C  =  0 ; 

and  with  the  x2  term  present,  Ax^x  replaces  in  the  above  ex- 
pression the  term 


11.    Illustrative  example.  —  Put  in  standard   form  and  plot 
carefully  ±y*-l2y  +  6x-n  =  0, 


completing  the  square  inside  the 
parenthesis. 


4a=-f;  a  =  -|. 


Plot  F,  ^,  .KFC',  and  the  further  point  where  y  =  %  ; 


1  0    _  8  .      rp  - 

- 


THE  PARABOLA  317 

Draw  a  smooth  curve  tangent  to  x  =  ±£-,  at  F(\°-,  f )  through 
the  points  which  are  plotted ;  here  it  would  be  well  to  find 
from  the  original  equation  the  intercepts  on  the  axes. 

PROBLEMS 
Put  in  standard  form  and  plot : 

1.  4#2—  12 y  +  6x  — 11  =0.      5.    (y  —  3)2  =  8 x  + 11. 

2.  4  x2  -  12  x  -6y  -11  =  0.      6.   yz  =  6x  +  ll. 

8.  Solve  graphically,  to  1  decimal  place, 

x2  +  y2  =  25 

yz  =  Sx,  by  drawing  both  graphs  to  the  same  axes. 

Put  in  standard  form  and  plot  the  equations  obtained  in  the 
two  following  problems : 

9.  The  formula  for  the  height  of  a  bullet  shot  vertically 
upward  with  a  velocity  of  800  feet  per  second,  s  =  800 1  —  16 12. 

10.  The  formula  for  the  time  of  beat,  in  seconds,  of  a  pendu- 
lum is  tz  =  —  -I,  taking  g  =  980  and  /  measured  in  centimeters  ; 


9 
taking  g  =  32,  /  must  be  measured  in  feet. 

Compute  corresponding  values  of  t  and  I  by  logarithms,  cor- 
rect to  3  significant  figures.  Would  the  diagram  be  changed 
if  g  is  taken  as  982  instead  of  980  ?  At  sea  level  on  the  equator 
g  =  978.1  cm./sec.2 ;  at  Washington,  980.0 ;  at  New  York,  980.2 ; 
at  London,  981.2  ;  or  in  feet/sec.2  32.09, 32.15,  32.16,  and  32.19, 
respectively. 

11.  The  Hell-Gate  steel  arch  bridge  in  New  York  is  one  of 
the  largest  arch  bridges  in  the  world.  See  the  illustration, 
p.  353.  The  lower  arc  of  the  arch  is  a  parabola,  977.5  feet  as 
span  and  220  feet  as  height  of  the  arch.  Write  the  equation 
of  the  arc,  taking  as  ic-axis  the  tangent  at  the  vertex  of  the 
parabola  and  as  y-axis  the  axis  of  the  parabola.  Compute 
4  a  to  1  decimal  place.  The  roadway  is  130  feet  above  the 


318  UNIFIED  MATHEMATICS 

base  of  the  arch  ;  compute  the  length  of  the  roadway  between 
the  parabolic  arcs.  There  are  23  panels  or  openings,  spaced 
42.5  feet  apart  at  the  centers ;  compute  the  vertical  lengths  to 
the  roadway  from  the  arc  of  the  parabola,  also  to  one  decimal 
place.  Compute  the  approximate  length  of  the  parabolic  arch  it- 
self by  computing  the  lengths  of  the  twenty-three  chords  on  the 
parabola ;  note  that  only  12  computations  are  necessary  ;  do  not 
carry  beyond  tenths  of  a  foot,  as  hundredths  would  have  little 
significance.  Locate  the  focus  and  the  directrix  of  this  parabola. 

12.  Engineers  use  the  following  method  for  constructing  a 
parabolic  arch ;    show  that  it  is  correct.     Suppose  that  it  is 
desired  to  construct  a  parabolic  arch  of  width  100  feet  and 
height  30  feet ;  a  rectangle  50  by  30  is  drawn  and  the  right- 
hand  side  is  divided  into  10  (or  n)  equal  parts  which  are  joined 
to  the  upper  left-hand  vertex  of  the  rectangle  (and  parabola) 
by  radiating  lines  ;  the  upper  horizontal  side  is  also  divided 
into  10  (or  n)  equal  parts  and  ordinates  are  drawn  at  these 
points  ;  corresponding  lines  intersect  at  points  on  the  parabolic 
arch  desired.     Of  the  lines  drawn  the  second  (or  rth)  ordinate 
to  the  right  of  the  vertex  corresponds  to  the  second  (or  rth) 
radiating  line  drawn  from  the  vertex  to  the  second  (or  rth) 
point  of  division  from  the  top,  on  the  right-hand  side. 

13.  If  an  ordinary  automobile  headlight  reflector  is  cut  by 
a  plane  through  its  axis  the  section  is  a  parabola  having  the 
light  center  as  focus.     If  the  dimensions  of  the  headlight  are 
10  inches  in  diameter  by  8  inches  deep,  locate  the  focus. 

14.  Locate  the  focus  of  a  parabolic  reflector,  6  inches  in 
diameter  and  4  inches  deep  ;  5  inches  deep  ;  6  inches  deep. 

15.  The  cable  of  a  suspension  bridge  whose  total  weight  is 
uniformly  distributed  over  the  length  of  the  bridge  takes  the 
form  of  a  parabola.     Assuming  that  the  cable  of  the  Brooklyn 
bridge  is  a  parabola,  which  it  is  approximately,  find  the  equa- 
tion in  simplest  form ;  the  width  between   cable   suspension 
points  is  about  1500  feet  and  the  vertex  of  the  curve  is  140  feet, 
approximately,  below  the  suspension  points. 


THE  PARABOLA  319 

16.  The  Kornhaus  Bridge  over  the  Aar  at  Berne,  Switzer- 
land, has  for  central  arch  a  parabola ;  the  span  is  384  feet  and 
the  height  of  the   arch   is   104.  feet.     If   there   are   vertical 
columns  spaced  24  feet  apart,  determine  the  length  of  these 
columns,  assuming  that  the  roadbed  is  30  feet  above  the  vertex 
of  the  parabola.     If  the  floor  of  the  roadbed  is  on  a  2.7  per 
cent  grade,  determine  the  difference  in  elevation  between  the 
center  of  the  bridge  and  the  ends. 

17.  The   parabolic    reflector    at    the   Detroit   Observatory, 
University  of  Michigan,  has  a  diameter,  which  corresponds  to 
arch  span,  of  37.5  inches;  the  focal  length  of  the  mirror  is 
19.1  feet,  from  vertex  to  focus.     Determine  the  height  of  the 
arch  (or  the  depth  of  the  reflector) ;   determine  the  equation 
of  the  parabolic  curve.     The  rays  from  a  sun  or  star  which 
strike  this  surface  parallel  to  the  axis  of  the  parabola  converge 
at  the  focus.     What  is  the  slope  of  this  mirror  at  the  upper 
point  of  the  mirror  ?     At  the  point  whose  abscissa  is  ^  inch  ? 

18.  To  draw  a  tangent  to  a  parabola  from  an  external  point 
you  can  proceed  as  follows  :  take  the  external  point  as  center, 
the  focal  distance  as  radius,  and  describe  an  arc  cutting  the 
directrix ;  from  the  point  of  intersection  draw  a  line  parallel 
to  the  axis ;  the  intersection  point  with  the  parabola  is  the 
point  of  tangency.     Prove  this  method. 

19.  Find  the  tangents  of  slope  +  \  and  —  .3  to  each  of  the 
parabolas  in  exercises  1  to  7 ;  time  yourself. 

20.  Find  the  tangent  to  each  of  the  parabolas  in  exercises 
1  to  7  at  the  point  on  each  parabola  whose  abscissa  is   +  2  ; 
the  exercise  should  be  completed  within  thirty  minutes.     Show 
the  geometrical  method  of  working  one  of  these  problems. 

21.  Find  the  tangents  to  the  parabola  in  problem  1  from 
the  point  (2,  10)  outside  the  parabola.     Describe  a  geometrical 
method  of  working  this  problem  after  the  graph  of  the  parab- 
ola is  drawn. 

22.  Plot  to  scale  with  the  dimensions  given  the  fundamental 
parabola  of  the  Alexander  III  Bridge. 


CHAPTER  XX 

THE   HYPERBOLA 

1.    Definition   and  derivation  of  the   equation.  —  (See  ellipse, 
page  289,  and  parabola,  page  309.) 

PF=e-PZ,         e>l. 

Take  FX  perpendicular  to  D'D  as  the  avaxis,  intersecting 
the  given  directrix  at  Q. 


Let  A  and  A'   divide  the  segment  QF  internally  and  ex- 
ternally in  the  ratio  e  (f  in  the  figure).     The   mid-point   of 

320 


THE  HYPERBOLA  321 

AA,  0  is  taken  as  the  origin  and  the  perpendicular  through 
this  point  to  X'X  as  the  y-axis,  OA  =  OA'  =•  a. 
Precisely  as  in  the  ellipse, 

AF=  e  •  AQ, 
A'F=e-A'Q; 

whence  AF  +  A'F  =  e  •  (AA'). 

AF+OA'+OF=2ae,  ' 
2  OF=2ae, 
OF=  ae, 


and,  by  subtraction, 


e 


F  is  (ae,  0)  ;  D'D  is  x  -  -  =  0. 

o 

The  relation  PF—  e  •  PZ  gives  the  equation, 


V(aj  -  ae)2  +  y* 

x*  +  a2e2  +  y2  =  e2z2  +  a2. 
x2(l  -  e2)  +  ?y2  =  a2(l  -  e2). 

Up  to  this  point  the  work  is  practically  identical  with  the 
work  in  the  case  of  the  ellipse  ;  here,  however,  1  —  e2  is  nega- 
tive, since  e  >  1.  Hence  we  write  this  equation, 


Let  62  =  a2(e2-l). 

/1-2         9/2 

5  __  .v_  _.  ^ 

a2     62 


2.    Geometrical    properties    of    the    hyperbola,    —  —     =  1.  — 

Since  y  and  —  ?/  lead  to  the  same  values  of  x,  the  curve  is 
symmetrical  with  respect  to  the  £-axis  ;  since  x  and  —  x  lead 
to  the  same  values  of  y,  the  curve  is  symmetrical  with  respect 


322 


UNIFIED  MATHEMATICS 


to  the  y-axis.     Hence  the  intersection  of  the  two  axes  is  the 
center  of  the  curve. 

6 

a 


Solving  for  y, 


this  expression  shows  that  the  curve  is  symmetrical  with 
respect  to  the  ic-axis,  for  any  value  of  x  gives  two  values  of  y 
equal  in  value  but  opposite  in  sign.  Since  any  value  of  x 


Symmetry  of  the  hyperbola 

numerically  less  than  a  gives  imaginary  values  of  y,  the  curve 
lies  wholly  outside  the  region  bounded  by  x  +  a  =  0  and 
x  —  a  =  0.  When  x  =  ±  a,  y  =  0  ;  these  are  self-corresponding 
points  on  the  horizontal  axis  of  symmetry  ;  these  points  are 
called  the  vertices.  When  y  =  0,  x  is  imaginary  ;  the  vertical 
axis  of  symmetry  does  not  intersect  the  curve. 


Solving   for  x,  x  =  ±  - 
6 


+  &2  ;  the  curve  is  symmetrical 

with  respect  to  the  y-axis  ;  every  real  value  of  y  gives  two 
corresponding   real   values   of   x,  symmetrically  placed  with 


THE  HYPERBOLA  323 

respect  to  the  y-axis ;  the  vertical  axis  of  symmetry  does  not 
cut  the  curve  in  real  points.  As  y  increases  in  value,  without 
limit,  so  do  the  two  corresponding  values  of  x  increase  in 
value,  numerically  without  limit. 

Since  there  is  this  vertical  axis  of  symmetry  it  is  evident, 
precisely  as  in  the  ellipse,  that  there  is  a  second  focus, 

F2(—  ae,  0),  and  a  corresponding  directrix,  x  +  -  =  0. 

e 

The  axis  which  cuts  the  curve  is  called  the  principal  axis ; 
the  other  axis  is  called  the  conjugate  axis.  The  lines  x  =  ±  a 
are  tangent  to  the  curve  at  the  vertices.  See  page  310. 

3.   Right  focal  chords.  —  The  foci  are  the  points  (ae,  0)  and 


(—  ae,  0)  ;    when  x  =  ±  ae,   y=±-  Va2e2  —  a2  =  ±  b  Ve2  —  1 ; 

a 

62 

but  since  62  =  a2(e2  — 1),  these   values  of  y  equal   ±  — ;  each 

a 

2  62 
right   focal   chord   is  of  length  -  — ,  and  is   constructed   by 

erecting  at  the  focus  lines  perpendicular  to  the  principal,  or 

transverse,  axis  of  length  —  on  each  side  of  the  axis. 

a 

The   foci  are  at   a  distance    ±  ae  from   the   center ;   now 


62  =  a2(e2  —  1)  gives  ae  =  Va2  +  b2,  which  is  the  length  of  the 
diagonal  of  a  rectangle  of  sides  a  and  6. 


4.   Finite   and   infinitely  distant   points   on   the  hyperbola.  — 

yJ2  ii'l 

Given   to  plot  the  hyperbola  -•  --  !j—  =  1,  note  that  a2  =  16, 

16     49 

&2  =  49;    49  =  16(e2-l),    whence    e2-l=±f,    e2  =  f|,    and 


It  will  be  found  convenient  to  draw  the  rectangle  having 
0  as  center  and  extending  4  units  to  the  right  and  left  of  0 
and  7  units  above  and  below.  The  half-diagonal  of  this 


324 


UNIFIED  MATHEMATICS 


rectangle  has  the  length  V65,  and  may  be  used  to  locate  on 
the  principal  axis  the  two  foci. 


X          U 

Asymptotes  and  one  branch  of  the  hyperbola —  =  1 

16      49 

The  line  y  =  x  cuts  the  curve  at  —-  —  =  1,     33a;2    =1; 

16     49        '  16  x  49 

=  ±  4.88. 


V33  33 

The  line  y  =  mx   cuts  the  curve  in  two  points,  whose  ab- 
scissas are  given  by 

49— 16m2''  V49  -  16  m2 


16       49 


THE  HYPERBOLA  325 

As  16  m2  approaches  nearer  and  nearer  to  49  these  two 
points  of  intersection  move  farther  and  farther  off;  when 
49  —  16  m2  =  0,  m  =  ±  |-,  the  two  points  of  intersection  of  each 
of  these  lines  with  the  hyperbola  move  off  to  an  infinite  dis- 
tance ;  the  lines  y  =  |-  x  and  y  =  —  ^  x  are  called  asymptotes  of 
this  hyperbola,  intersecting  the  curve  in  two  coincident  points 
both  at  an  infinite  distance. 

x^      w2  b 

In   the   hyperbola    - —  —  =  1.   the  two    lines   y  =  —  x  and 
a2      &2  a 

y  = x,  or  -  —  9-  =  0  and    -  +  ^  =  0  are  asymptotes  :   note 

a  a     6  a     b 

that  (-  —  ^  l  ( -  +  - 1  gives  the  left-hand  member  of  the  equa- 
\a     bj  \a     bj 

tion  of  the  hyperbola,  when  the  right  hand  is  unity. 
5.    Illustrative  problems.  —  Plot  the  hyperbola 

^2__^  =  1 

16     49 

The  rectangle  of  sides  8  and  14,  parallel  to  x-  and  y-axes  respectively, 
is  plotted  with  its  center  at  the  origin.  As  noted  above,  the  diagonals 
of  this  rectangle  give  the  distance  from  the  center  on  the  transverse  axis, 
here  horizontal,  of  the  foci ;  the  diagonals  extended  are  the  asymptotes, 
and  to  these  lines  the  curve  approaches  more  and  more  nearly  as  the 
curve  recedes  towards  infinity.  The  right  focal  chords  have  the  total 

2  ft2  49 

length  — ;  plot  —  vertically  above  and  below  the  foci.     Take  x  =  6, 

a  4 

this  gives  another  point  between  vertex  and  right  focal  chord  ; 

S  =  ^"15  y  =  ±7vT25  =  ±  7(1.12)  =±7.84; 
49      lo 

take  x  =  10,  y  =  ±  7  v/OS  =  ±  7(2.29)  =  ±  16.03. 

Plot  also  the  symmetrical  points. 

PROBLEMS 

1.  Plot  the  hyperbola  —  -  £-  =  1. 

49     16 

2.  Plot  the  hyperbola  — y—  =  1. 

100     100 

This  type  of  hyperbola,  a  =  6,  is  called  an  equilateral  or  rec- 
tangular hyperbola.  Why  ? 


326 


UNIFIED   MATHEMATICS 


3.  Plot  the  hyperbola  ^-—-^-  =  1,     computing  values   re- 

o7      59 

quired  to  one  decimal  place. 

4.  The  equation  of  the  hyperbola  —  —  ^-  =  1  may  be  put  in 


parametric  form, 


Noting  that  sec  0  is 


x  =  a  sec  0 
y  =  b  tan  0 

1 


cos  6 


,  find   by  using  logarithms    five 


points  011  each  of  the  above  hyperbolas.  Check  on  the  graphs 
drawn.  What  is  the  geometrical  significance  of  0  ? 

5.  Find  the  equations  of  the  asymptotes  of  each  of  the  pre- 
ceding hyperbolas  and  find  to  1'  the  angle  of  inclination  to 
the  horizontal  axis. 

6.  Compare    the     right-hand     branch    of     the     hyperbola 

x*      v2  81 

—  —  ~  =  1,    with  the  parabola  ?/2  =  —  (x  —  4) ;  this  parabola 

J-O      y  lo 

has  the  same  vertex  and  passes  through  the  extremities  of  the 
right  focal  chord.  Prove  this.  Do  these  curves  coincide  in 
other  points  ? 


m 


i&r, 


Focal  distances  to  Pi  Ui,  y\)  on  the  hyperbola 


THE  HYPERBOLA 


327 


6.  Difference  of  the  focal  distances  constant.  —  Designate  the 
right  focus  by  FI  and  the  corresponding  directrix  by  DD', 
and  the  left  focus  by  FZ)  having  D^Df»  as  the  corresponding 
directrix.      Then  the  focal  distances  PFl  and  PF2  are  ex^  —  a 
and   exi  +  a,   respectively  ;    the   difference,   PFZ  —  PF^  =  2  a, 
is  constant. 

The  hyperbola  may  be  defined  as  a  curve  generated  by  a 
point  which  moves  so  that  the  difference  of  its  distances  from 
two  fixed  points  is  constant. 

7.  Standard  forms  of  the  equation  of  the  hyperbola.  — 

(«  -  /Q*  _  (y  —  *02  _  i 

A  TO 

a2  &2 

(y  _  fc)2  _  (g  —  A)2  _   . 

^^  62 

Precisely  as  in  the  ellipse,  article  6,  Chapter  18,  the  equation 

of  the  hyperbola  —  —  |^  =  1,  may  be  interpreted 

Of          0 


OM2     MP2 


-i 


For  any  hyperbola  whose  axes  of  symmetry  are  parallel  to  the 
coordinate  axes  we  obtain,  from  this  relation,  the  equations 


a2  62  and  a2  62 

for  a  horizontal  hyperbola,  for  a  vertical  hyperbola. 


Horizontal  hyperbola 


3s »  I     4-  - 


•m 


Vertical  hyperbola 


328  UNIFIED  MATHEMATICS 

8.    Conjugate  hyperbolas  and  limiting  forms  of  the  hyperbola 
equation.  —  Given  any  hyperbola, 

X2_y2=1 

a2     62 

the  lines 

a 

represent  the  asymptotes  ;  the  equation 


a2      &2 

represents  a  vertical  hyperbola  about  the  same  rectangle  and 
having  the  same  asymptotes.     Any  two  hyperbolas  so  related 
are  called  conjugate  hyperbolas. 
Illustration. 

(z-3)2     (y  +  2)2  ,  (z-3)2     (y  +  2)2 

—  '-  =  1  and  *  —  —  4  =  —  1  are  con- 

16  49  16  49 

jugate  hyperbolas  ;   the  second  is  written  in  standard  form 

**-~  —  i  —  L^  —  '-  =  1,  wherein  the  focal  distances  ±  ae  from 
49  16 

the  center,  and  the  distances  ±  -  of  the  directrices  from  the 

e 
center  (3,  —  2)  are  obtained,  regarding  o2  as  49  and  i2  as  16. 

The  asymptotes  of  these  conjugate  hyperbolas  are  given  by 
the  equation   i—  —  —  -  —  isLJI  —  L  =  0,  or  by  the  equivalent  in 

JLO  ^rt/ 

separate  factors, 


7 

fy>  /A2  (y    fc\2 

The  equation  ^ *• —  *t '-  =  0,  representing  two  real 

straight  lines,  is  a  limiting  form  of  the  hyperbola  equation, 

(x  -  h)2      (y  +  fc)2  b 

— -t-  =  m.     Asm  approaches  zero,  -  remains  con- 
a2  b2  a 

stant  and  the  hyperbola  approaches  more  and  more  nearly  the 

•  •L.T         x  —  h      11  —  k     n      j  x  —  h  .  y  — k      n 
two  straight  lines =  0  and 1-  * :  =  0. 


THE  HYPERBOLA 


329 


9.  The  equilateral  or  rectangu- 
lar hyperbola.  —  The  hyperbola 

2«2  y2 

-—  =  1  is  called  an  equilat- 

a2     a2 

eral  hyperbola  since  b  =  a  ;  it  is 
also  called  a  rectangular  hyper- 
bola as  the  asymptotes  are  at 
right  angles  to  each  other. 

Since  62  =  a2  (e2  —  1),  the  value 
of  e  in  an  equilateral  hyperbola 
is  V2  or  1.414;  for  e  >V2, 
52  >  a2 ;  for  e  <  V2,  &2  <  a2. 


X£;0- 

HfC 

'A  :  . 


The   equilateral   or  rectangular 
hyperbola 


10.    Illustrative  problem.  —  Put  the  equation 

4  x2  +  16  x  -  9  y*  -  18  y  -  75  =  0 
in  standard  form  and  plot  the  curve. 

)  =  75 
4x  +  4)-  9(i/2  +  2y  +  1)=  75  +  16  -  9 


20.5  9.11 

The  center  is  at  (—2,  —1);    the  hyperbola  is  of  horizontal  type; 
a2  =  20.5  and  a  =  4.53  ;  62  =  9.11  and  b  =  3.02  ; 


ae  =  V20.5  +  9.11  =  V29761  =  5.44  ;   —  =  —  ; 

a      4.53 


Sari 


further  convenient  points 
are  given  by  x  =  5  ;  substi- 
tuting in  the  original  is 
easiest,  giving 

9  y2  +  18  y  -  105  =  0  ; 


y=-l  ±  Vl 


Upper   portion  of  right-hand  branch  of  the 
given  hyperbola 


=  -l±3.56. 


330  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  Put  the  equation 

4  x-2  +  16  x  -  9  y*  -  18  y  +  107  =  0, 
in  standard  form  and  plot. 

2.  Plot  one  quarter  of  the  hyperbola 


1472      592 

3.  Plot  a  hyperbolic  arch,  width  200  feet,  height  60  feet, 
as  part  of  a   rectangular   hyperbola.     Assume   the  equation 
2/2  _  xz  —  a2}  an(j  note  that  (100,  a'+  60)  is  on  the  curve. 

4.  What  limitation  is  there  upon  the  values  of  A  and  B, 
if  the  equation  Ax2  +  By2  +  2  Gx  +  2  Fy  +  C  =  0  represents  a 
hyperbola  ? 

5.  Any  equation  of  the  form  xy  =  k 
or                   (a-p  +  b$  +  GI)  (a&  +  b$  -f  Cj)  =  k, 

has  for  its  locus  a  hyperbola  ;  the  lines  obtained  by  equating 
the  left-hand  member  to  zero  are  the  asymptotes.  Plot  the 
hyperbolas  xy  =  10  and  (x  —  3  y)(x  —  4  y)  =  50. 

6.  Put  the  following  equations  in  standard  form,  completing 
the  square  first  and  reducing  to  standard  form  by  division. 

a. 
b. 
c. 

d.  5  a2  -  12  y2-  117  =  0. 

e.  3  x2  -  24  z-4y2-  16^-52  =  0. 

7.  Plot  the  preceding  five  hyperbolas,  choosing  an  appro- 
priate  scale.     Plot   the  extremities   of   the   conjugate   axes  ; 
plot  the  rectangle  and  its  diagonals  ;  plot  the  extremities  of 
the    right    focal    chords  ;    plot    at   least    one    further    point, 
properly   chosen    to    give   the    form    of   the    curve,   and   its 
symmetrical  points  with  respect  to  the  axes.     It  is  desirable 
to  plot  at  least  two  of  these  curves  completely  ;  the  remaining 
curves  need  be  sketched  only  in  the  first  quadrant. 


THE  HYPERBOLA  331 

8.  Determine  a2  and  62  to  one  decimal  place  in  the  follow- 
ing three  hyperbolas  : 

a.   17  x2  -  43  y2  =  397. 

6.    5  a;2  -  17  x  -  10  ?/2  -  35  y  =  0. 

c.    7(«  -  2)2  -  B(y  -  3)2  =  39. 

9.  In  the  three  hyperbolas  immediately  preceding  deter- 

fi2  CL 

mine  ae,  —  ,  and  -  to  one  decimal  place. 
a  e 

10.  In  each  hyperbola  of  problem  8  determine  x  when  y  =  2. 

11.  Using  the  data  of  the  three  preceding  problems,  plot 
the  three  hyperbolas  of  problem  8. 

/v.2  y2 

12.  In  the  hyperbola  --  *—  =  1,  find  the  coordinates  of  the 

64     36 

foci.  What  is  the  distance  of  the  point  whose  abscissa  is  12 
from  each  of  the  foci  ?  of  the  points  whose  abscissas  are  10,  11, 
15  ?  State  the  general  form  for  this  distance. 

13.  Put  the  following  equations  in  standard  form  and  discuss 
the  curves  represented  by  these  equations  : 

a.   cc2  —  6  x  -  y2  —  6  y  =  0. 
6.    x2-6a-4/2- 


36 


CHAPTER   XXI 

TANGENTS   AND   NORMALS   TO    SECOND    DEGREE 
CURVES 

1.  The  general  quadratic  in  x  and  y.  —  The  general  equation 
of  the  second  degree  in  x  and  y  is  written, 

Ax2  +  2  Hxy  +  By2  +  2  Gx  +  2  Fy  +  C  ••=  0. 

The  equations  of  the  circle,  parabola,  ellipse,  and  hyperbola 
are  special  types  of  this  general  equation.  Since  none  of 
these  standard  forms,  representing  curves  of  the  second  degree 
with  axes  of  symmetry  parallel  to  the  coordinate  axes,  have 
an  xy  term,  we  find  it  convenient  to  discuss  the  general 
equation,  with  H=0,  or 

Ax*  -f  By2  +  2  Gx  -f  2  Fy  +  C  =  0. 

This  represents  one  of  the  curves  —  circle,  ellipse,  parabola, 
or  hyperbola  —  mentioned  above,  or  some  limiting  form  of  the 
same,  including  pairs  of  lines  and  imaginary  types.  It  can  be 
shown  that 

Ax2  +  2  Hxy  +  By*  +  2Gx  +  2Fy  +  C=0 

represents  no  new  curve ;  simply  one  of  the  above-mentioned 
types  turned  at  an  angle  to  the  coordinate  axes. 

2.  General  equation  of  the   second   degree   represents  a  conic 
section.  —  Given  a  right  circular  cone,  it  can  be  shown  by  the 
geometrical  methods  of  Euclidean  geometry,  that  the  section 
which  is  made  with  the  surface  of  the  cone  by  any  plane  is 
one  of  the  curves  above  mentioned ;  thus  a  plane  parallel  to 
the  base  cuts  the  cone  in  a  circle,  or  in  a  point  circle  if  through 
the  vertex. 

332 


TANGENTS  AND  NORMALS  333 

The  cone  is  conceived  as  the  whole  surface  determined  by 
the  straight  line  elements  of  the  cone  produced  to  infinity. 

A  plane  which  runs  parallel  to  only  one  of  the  elements  cuts 
the  cone  in  a  parabola,  or  in  two  coincident  lines  if  the  plane 
passes  through  an  element  and  a  tangent  to  the  circular  base 
of  the  cone. 

A  plane  which  cuts  all  the .  elements  in  finite  points  cuts 
the  cone  in  an  ellipse ;  this  is  a  point  ellipse  when  the  plane 
passes  through  the  vertex  of  the  cone. 

A  plane  which  cuts  the  cone  parallel  to  the  plane  of  two 
elements  cuts  it  in  a  hyperbola ;  if  the  plane  passes  through 
the  vertex  the  hyperbola  reduces  to  two  straight  lines. 

3.  Historical  note  on  conic  sections.  —  The  fundamental  prop- 
erties of  conic  sections  were  discovered  by  Greek  mathema- 
ticians nearly  two  thousand  years  before  the  invention  of 
analytical  geometry  which  was  perfected  by  Descartes  and 
Fermat,  French  mathematicians  of  the  seventeenth  century. 
A  treatise  on  conies  was  written  by  Euclid  (c.  320  B.C.),  but  it 
was  entirely  superseded  a  century  later  by  a  treatise  by  Apol- 
lonius  (c.  250  B.C.)  of  Perga,  whose  treatise  included  most  of 
the  fundamental  properties  which  we  discuss.  The  proper- 
ties of  the  parabola  connected  directly  with  focus  and  directrix 
are  not  included  in  the  eight  books  (chapters)  on  conic  sec- 
tions by  Apollonius,  nor  was  the  directrix  of  the  central  conies 
employed  by  him.  Pappus  of  Alexandria  (c.  300  A.D.),  almost 
the  last  of  the  Greek  mathematicians  of  any  note,  included 
these  in  his  Mathematical  Collections. 

The  Greek  mathematicians  were  interested  in  these  curves 
for  the  pure  geometrical  reasoning  involved.  That  the  paths 
of  the  planets  were  conies  they  did  not  know ;  nor  did  they 
know  any  practical  applications  of  these  conies.  However, 
the  fact  that  Greek  mathematicians  had  studied  these  prop- 
erties made  it  possible  for  John  Kepler  and  Isaac  Newton 
to  establish  the  laws  of  movement  of  the  planets  in  the  uni- 
verse in  which  we  live.  The  men  mentioned  and  Nicolas 


334  UNIFIED  MATHEMATICS 

* 

Copernicus,  who  reasserted  the  heliocentric  theory  of  the  uni- 
verse, were  all  thoroughly  versed  in  the  pure  geometry  of  the 
Greeks ;  their  new  theories  were  built  directly  upon  this 
foundation  of  pure  geometry. 

4.  Tangent  of  slope  m  to  a  second  degree  curve.  —  Any  line 
y  =  mx  -f  k  cuts  a  curve  given  by  an  equation  of  the  second 
degree  in  two  real  points,  or  in  two  imaginary  points,  or  in 
two  coincident  points.  The  abscissas  of  the  points  of  inter- 
section are  given  by  the  quadratic  in  x  obtained  by  substitut- 
ing y  =  mx  -+-  k  in  the  equation  of  the  curve  ;  the  two  equations 
are  solved  as  simultaneous  equations.  The  condition  for  tan- 
gency  is  that  the  two  points  of  intersection  of  the  line  with 
the  curve  shall  be  coincident ;  this  will  be  the  case  when  the 
two  roots  of  the  quadratic  in  x,  i.e.  the  two  values  of  the 
abscissas  of  the  points  of  intersection,  are  equal. 

When  the  intersections  of  a  line  with  a  quadratic  curve  are 
given  by  a  linear,  instead  of  a  quadratic,  equation,  the  mean- 
ing is  that  one  point  of  intersection  has  moved  off  to  an  in- 
finite distance.  As  the  coefficient  of  the  square  term  of  a 
quadratic  approaches  zero  one  root  becomes  larger  and  larger 
without  limit ;  see  page  98. 

Parabola,       y2  =  4  ax, 

y  =  mx  +  k. 

Solving,          a.==-(fcm-2a)±V4a(q-7nfc)> 

m2 

For  equal  roots,  or  coincident  points,  k  =  — . 

m 

,   a  •  .  f  a     2  a\ 

A.   y  =  mx  -\ —  is  tangent  to  w2  =  4  ax  at  [  — ,  -  -  i« 
m  \m?     m  J 

yS.  y2 

Ellipse,  — (-  —  =  1,  and  the  line  y  =  mx  -f  k. 
a2     62 


Solving, 


TANGENTS  AND  NORMALS  335 


B.   y  =  mx  ±  -\/a?m2  -f-  62  is  tangent  to  the  ellipse 


^  I  y2  =  1  at  x  =  T  a'm  Va2m2  +        v== 


a2     62  a2m2  +  62  Va2m2  +  62 

For  every  value  of  m  there  are  two  real   tangents  to  an 
ellipse.     Similarly 


C.  y  =  mx±  Va2m2  — 

is  tangent  to  the  hyperbola, 


a2     b2  Va2m2  -  6s 

For  values   of   |  m  |  >  - ,  there  are  two  real  tangents  to  a 
a 

hyperbola ;  for  |  m  |  <  -  the  tangents  are  imaginary ;  for 
m  =  ±  - ,  there  is  only  one  tangent  and  its  point  of  tangency 

d 

is  at  an  infinite  distance,  or,  as  noted  before,  the  lines  y  =  ±  -  x 

a 

are  asymptotes  of  the  curve. 

The  method  of  this  article  is  employed  in  deriving  tangents ; 
the  equations  given  under  A,  B,  and  C  above  are  used  mainly 
in  proving  geometrical  properties  of  these  curves.  Note  that 
if  these  equations  are  used  as  formulas,  they  apply  only  to 

curves  of  the  type  given;   y  =  mx  -\ —  gives  the  tangent  only 

m 

to  a  parabola  of  the  form  y2  =  4  ax  (a  may  be  positive  or  nega- 
tive). Similarly,  y  —  mx±  Va2m2  +  &2  gives  the  tangent  of 
slope  m  only  to  the  ellipse, 

—  +  y~  =  1,  (a2  may  be  less  than  ft2). 


336  UNIFIED  MATHEMATICS 

PROBLEMS 

Find  tangents  of  slope  2  and  of  slope  —  3  to  the  following 
three  curves.     Follow  the  method  of  article  4. 

1.  x2  +  y2  —  10  x  =  0 ;  find  the  points  of  tangency. 

2.  3y2  —  4  a;  —  6  y  =  0 ;  find  also  the  normal  of  slope  —  ^. 

3.  x2  +  3y2  —  4ce  —  6y  =  0  ;    find  the  diameter  joining   the 
two  points  of  tangency. 

4.  xy  —  25—0.     Find   the  tangents  of   slope  —  2,  and  the 
points  of  tangency.     Find  the  tangent  of  slope  m.     For  what 
values  of  m  are  the  tangents  imaginary  ?     Plot  10  points  on 
this  curve.     Where  do  all  points  of  this  curve  lie  ?     Find  the 
intersections  of  y  =  .01  x  -j-  5  with  this  curve. 

5.  Find  the  tangents  at  the  extremities  of  the  right  focal 
chord   of  yz  =  8  x ;    where   do  they   intersect  ?     Similarly  in 
7/2  =  4  ax.     Is  this  true  of  any  parabola  ?     Explain. 

6.  Find   the   tangents   at   the   extremities   of   either   right 
focal  chord  of  —  +  —  =  1 ; 

—  '  *  <J 

where  do  they  intersect  ?     Similarly  with 

S  +  fh1'^'^} 

Where  do  these  tangents  intersect  ?     What  change  is  necessary 
to  prove  this  property  for  the  hyperbola  ?     Explain. 

7.  Find  the  perpendicular  from  the  focus  of  y2  =  8  x  to  the 
tangent  of  slope  2 ;   where  do  they  intersect  ?     Similarly  for 
the  tangent  of  slope  m.     State  what  you   have   found   as   a 
property  of  any  parabola. 

8.  In  the  ellipse  —  +  *-  =  1,  find  the  perpendicular  from  the 

— • '      y 

focus  to  the  tangent  of  slope  2  ;  find  the  point  of  intersection ; 
note  that  it  is  a  point  on  the  circle,  x2  +  y2  =  25.     Prove  that 


TANGENTS  AND  NORMALS  337 

the  same  is  true  of  the  perpendicular  from  the  focus  upon 
any  tangent  of  slope  m.     Prove  that  in  the  ellipse 

xZ  _i_  yl — i 

a2      b- 

the  perpendicular  from  the  focus  upon  any  tangent  meets  it 
on  the  major  auxiliary  circle. 

9.  Follow  the  directions  of  problem  8  with  the  hyperbola, 

16~9=:1' 

2         «/2 

and  —  =  1. 

a2     62 

making  necessary  changes. 

10.  Find   the  angle  between  y  =  5x  —  7  and   the   parabola 

The  angle  between  a  straight  line  and  a  curve  is  defined  to 
be  the  angle  between  the  straight  line  and  a  tangent  to  the 
curve  at  the  point  of  intersection. 

NOTE.  —  Solve  for  the  points  of  intersection ;  write  the  equation  of 
tangent  at  each  point  of  tangency  ;  find  the  angle  between  each  tangent 
and  the  line  y  =  5  x  —  1.  See  article  8,  chapter  15.  Check  by  finding 
the  slope  angle  of  the  lines  y  =  5  x  —  1  and  of  the  tangent  lines. 

11.  Find  the  angle  between  y  =  2x  —  5  and  #2 -f  y2  =  100, 
at  each  point  of  intersection.     Check  as  in  problem  10. 

5.  Tangent  at  a  point  (x1}  |/j)  on  a  curve  given  by  an  algebraic 
equation.  —  On  any  curve  a  line  joining  a  point  P:  to  a  point 
P2  is  called  a  secant ;  obviously  this  secant  cuts  the  curve  in 
two  distinct  points.  If  P2  approaches  Pl  along  the  curve,  the 
secant  changes,  approaching  more  and  more  nearly,  in  general, 
a  definite  limiting  line.  The  limiting  position  of  the  secant  is 
called  the  tangent  to  the  curve  at  P±. 

The  analytical  method  of  obtaining  this  limiting  line  is  as 
follows : 

Take  P^  fa,  y^)  any  point  on  the  curve ; 


338  UNIFIED  MATHEMATICS 


Take  P2  as  (xi  +  '*>  y\  +  &)  a^80  ou  the  curve  ;  the  chord  PiP2 

k 
has  the  equation  y  —  yl  =  -  (x  —  a^). 

k 
Find  the  value  of  -  conditioned  by  the  fact  that  P2  and  P: 

fv 

both  lie  on  the  curve,  by  substituting  fa,  y^)  and  (o^  +  h,yv  +  k) 
in  the  given  equation  and  subtracting,  member  for  mem- 
ber. 

k 

This  value  of  -  will  be  found,  in  general,  to  have  a  definite 
h 

limiting  value  as  A;  and  h  approach  zero  ;  this  limiting  value  is 
the  slope  of  the  tangent. 

The   method   outlined   applies   to   any  curve  given  by  an 
algebraic  equation. 

?/2  =  4  ax  •  P^X!,  2/j)  on  curve  ;  P2(ajj  +  7i,  yi  +  k)  on  curve  ; 
y  —  y\  =  j(x  —  *i)>  chord  joining  PtP2. 

(yi  +  *)2  =  4a(aj!  +  A),      or      y?  +  2  kyl  +  k2  =  4  ax1  +  4  ah, 
since  P2  is  on  curve. 

y^  =  4  axi,  since  Px  is  on  curve. 
-?/!  +  k2  =  4  a/i,  by  subtraction. 


-  =  -  -  —  gives  the  slope  of  the  chord  joining  Pt  to  P2. 
h     2  y-L  +  k 

k 
Let  h  approach  0,  k  also  approaches  0,  but  -  always  equals 

III 

and  this  value  approaches  more  and  more  nearly  to 


-  or  —  as  a  limit  :  this  limit  is  the  slope  of  the  tangent. 
2  yi       yi 

y  —  yi  =  —  (x  —  x1)  is  the  tangent  to 

y\ 

z  =  4:ax  Sit   x          on  curve. 


TANGENTS  AND  NORMALS  339 

This  equation  may  be  simplified, 

y$  •=  2  ax  +  ft2  —  2  aXi ;  but  ft2  =  4  ax,  whence 
fty  =  2  a(x  +  xt),  which  is  the  tangent  equation. 
By  precisely  this  method,  the  tangent  to 

x2      v2 

h  —  =  1>  at  (xi,  Vi) 

a2     62 

on  curve  has  been  found,  in  section  11,  chapter  18,  to  be 

~a?+~P~ 
The  tangent  to 

—  =  1,  at  (x,,  ft)  on  curve  is  — ^  =  1. 

a2     62  a2       &2 

The  tangent  to 

Ax2  +  By2  +  2  Gx -{-2  Fy  +  C=0,  at  (xl5  ft)  on  curve  is 

The  tangent  to 

^lx2  +  2  Hxy  +  Bif  +  2  Gx  4-  2  /fy  +  (7  =  0,  at  (x1?ft)  on  curve  is 

-G(x  +  x1)  +  ^  +  ft)+C'=0. 


All  the  preceding  are  embraced  in  the  last  formula,  as 
special  cases.  The  final  form  should  be  remembered  and  used 
as  a  formula. 

The  above  special  forms  for  the  equations  of  the  tangents  to 
conies  given  by  equations  in  standard  form  may  be  used  to 
derive  tangential  properties  of  these  curves.  Some  of  these 
properties  are  touched  upon  in  the  problems  below  and  will 
recur  in  the  next  chapter. 


340 


UNIFIED  MATHEMATICS 


6.    Tangential  properties  of  the  parabola.  — 

y*  =  4  ax,  any  parabola  Pv  VP2- 

y^  =  2  a(x  -}-  Xi),  tangent  at  P\(xl}  y^  on  curve,  cutting  the 
axis  at  T,  the  directrix  at  Q,  the  vertex  tangent  at  S. 

y  —  y1  = p.  (£  _  Xl),  normal  PLN,  cutting  the  axis  at  N. 


m 

— -\t 


Tangential  properties  of  the  parabola 

T  and  ^V  are  obtained  as  the   ^-intercept   of  tangent   and 
normal,  respectively; 

T  is  (-  aj1}  0)  ;  N  is  (^  +  2  a,  0). 

Hence  FT  =  VM,  the  tangent  cuts  off  from  the  vertex  on 
the  axis  the  same  distance  that  the  perpendicular  from  the 
point  of  tangency  on  the  axis  cuts  off  from  the  vertex.  This 
gives  a  simple  method  of  drawing  a  tangent  to  a  parabola  : 
drop  the  perpendicular  P\M  to  the  axis  ;  extend  the  axis  from 
the  vertex,  making  VT  =  VM,  the  length  of  the  intercept; 
the  tangent. 


TF=x1-\-a  = 


TANGENTS  AND  NORMALS  341 

Hence  Z  FTP1  =  Z  FPl  T,  base  angles  of  an  isosceles  A. 

Now  Z  FTPi  =  Z  TP^Z,  alternate  interior  angles  of 
parallel  lines,  etc.  Z  FPl T—/.  TPLZ,  i.e.  the  tangent 
bisects  the  angle  between  a  focal  chord  and  a  line  parallel  to 
the  axis ;  the  normal  PiN  bisects  the  supplementary  angle 
FP^,  making  Z  FP1N'=  Z  NP,R. 

Further  S  is  the  mid-point  of  TPl  (since  VS  Y  is  parallel  to 
PiJ/and  bisects  the  side  TM). 

.-.  FS  is  perpendicular  to  TPl ;  the  perpendicular  from  the 
focus  to  any  tangent  meets  it  on  the  vertex  tangent. 

QF  being  drawn,  A  QFPl  =  A  QZP^ 
',  Hence  QFPl  is  a  right  angle. 

Extend  P^F  to  cut  the  parabola  at  P., ;  draw  P2Z2  to  the 
directrix,  and  P2Q. 

A  P2Z2Q  =  A  PZFQ. 

Hence  P2Q  bisects  the  angle  ZzPzF  &nd  is  the  tangent. 

Further  Z  P-2QPi  is  a  right  angle,  since  it  is  half  of  the 
straight  angle  about  Q. 

Summary  of  tangential  properties  of  the  parabola 

1.  The  tangent  bisects  the  angle  between  the  focal  radius 
and  a  line  parallel  to  the  axis ;   the  normal  bisects   the  in- 
scribed angle  between  the  focal  radius  and  a  line  parallel  to 
the  axis  through  the  point  of  tangency. 

2.  The  perpendicular  from  the  focus  of  any  parabola  on  any 
tangent  meets  it  on  the  vertex  tangent. 

3.  Tangents  at  the  extremity  of  a  focal  chord  meet  on  the 
directrix,  and  at  right  angles. 

4.  The  focal  chord  is  perpendicular  to  the  line  joining  the 
focus  to  the  intersection  on  the  directrix  of  the  tangents  at 
the  extremities  of  the  focal  chord. 


342 


UNIFIED  MATHEMATICS 


7.   Tangential  properties  of  the  ellipse  and  hyperbola.  — 

/**  '/"  /*"  iP" 

— |-  2-  =  1,  any  ellipse, ^-  =  1,  any  hyperbola. 

(JL          0  Ct         0 


a2        &2  ~                                  a*        b2 

tangent  at  fa,  y,)  on  curve,  cutting  the  principal  axis  in  th 
point  T. 

! 

^^~  

jr*-}*-  

._  !s--S  s  

__,  <f-^^  -,~~-  - 

^sj^-v. 

•+•                  :::::::::::::;;; 

'7)\i 

__    s    -t- 

"X,                               v-(  L          •  •'  ^         i  ' 

*,.         vii."J"       LrTl        j'/ii 

'   -  '-"              ^-        j  /    \ 

-—  ''^N.               ./          -    T/^  /-    / 

\  .  ! 

-T—  T  —  —  \'  •  ;;••• 

-r-i/i—  T  --  -.y-"—  - 

'  iv  "r-j-1              ^  SZ~      —    \   ~ 

i      .         / 

<_                                 _  S  ^_  __  ^ 

^T-±rr-j/---l' 

k_1  —  at_  i(':,  —  ^*aei^  - 

V 

•     •—   '  —--^    ......  f4-    — 

l_j_  i  .  ;  I  —  .1.1  —  i  —  —  —  ,  .  .  i  ^ 

KJ—  ^    '    !    '     ^  

=r: 

is  ^v     '  '     i 

lS^   ^^f 

T^Z      ,-_  —. 

^v^v^  —  ^"r" 

-^ 

r 

^^^ 

Tangential  properties  of  the  hyperbola 
The  tangent  bisects  the  angle  between  focal  radii  to  the  point  of  tangency. 


giving  the  normal  at  xt,  y^  on  curve. 


Tis    -,0j 


In  the  ellipse, 


,2r      A^ 

'  *^i?    /• 
In  the  hyperbola, 


=  —  -  ae;  F^N=  ae  -  &K*     F1T=ae  -  — ;  FlN=Ael— 


TANGENTS  AND  NORMALS  343 

The  lengths  i^JV  and  F2N  are  seen  to  be  proportional  to  the 
lengths  P^  and 


snce 


a  +  ex,.     ae  + 


ae 

Similarly      !    *  =  — *— ,  since  a  ~  e—  =  — 

v  T-»    T7t  TvT    nn  „.       I        ^  ^O 

^2^  a  +  ex,_     a2  . 


If  a  line  from  the  vertex  of  a  triangle  divides  the  opposite 
side  into  segments  proportional  to  the  adjacent  sides,  the  line 
bisects  the  angle  of  the  triangle ;  hence  the  tangent  and 
normal  at  any  point  on  the  ellipse  and  hyperbola  bisect  inter- 
nally and  externally  the  angle  between  the  two  focal  radii  to 
the  point. 

Another  method  of  constructing  the  tangent  is  to  construct 

a2 

— .     In  the  ellipse  the  major  auxiliary  circle   x2  +  y2  =  a2  is 

drawn  and  the  tangent  at  P3  (x1}  y2)  on  this  circle  x&  -f  ygf  =  a2 
has  the  intercept  — ;  draw  the  tangent  at  (xlt  y2)  to  the  circle, 

Xi 

cutting  the  X-axis  at  T ;  connect  T  with  Pt  on  the  ellipse. 

In  the  hyperbola  x±  >  a,  so  that  this  construction  cannot  be 
used ;  from  M(xl9  0)  on  the  X-axis  a  tangent  to  the  circle 

a:2  +  yz  =  a2  intersects  it  at  a  point  U  whose  abscissa  is  ^-,  since 

«j 

OT=  a 
a       x± 

Summary  of  tangential  properties  of  the  ellipse,  hyperbola,  and 
parabola,  regarding  the  parabola  as  having  a  second  focus  at 
an  infinite  distance  on  its  axis  of  symmetry. 

1.  The  tangent  to  an  ellipse,  hyperbola,  or  parabola  bisects 
the  angle  between  the  focal  radii  to  the  point  of  tangency. 

2.  The   perpendicular   from   the    focus   upon  any   tangent 
meets  it  on  the  circle  having  the  center  of  the  conic  as  center, 


344  UNIFIED  MATHEMATICS 

and  passing  through  the  principal  vertices.  In  the  parabola 
this  circle  has  an  infinite  radius  and  so  reduces  to  the  tangent 
line  at  the  vertex  of  the  parabola. 

3.  Tangents  at  the  extremities  of  a  focal  chord  meet  on  the 
directrix. 

4.  The  focal  chord  is  perpendicular  to  the  line  joining  the 
focus  to  the  intersection  on  the  directrix  of  tangents  at  the 
extremities  of  the  given  focal  chord. 

PROBLEMS 

1.  Find  the  tangent  to  the  curve  xy  =  25  at  the  point  (5,  5) 
by  the  method  of  article  5. 

2.  Find  the  tangent  to  x2  =  8  y  at  the  point  (5,  -2g5-)  by  the 
method  of  article  5. 

3.  Find  the  tangents  to  the  curves  in  problems  1-3  of  the 
preceding  set  of  problems  at  a  point  (x1}  yt)  on  each  curve  by 
the  general  formula. 

4.  Find    the    tangent    to    the    curve    x2  =  8  y  at   a   point 
(XD  y\)  on  the  curve ;  note  that  (x1}  y^)  satisfies  the  equation 
of  the  given  curve ;   find  a  second  equation  which  the  point 
(»!,  th)  must  satisfy  if  the  tangent  obtained  is  to  pass  through 
(5,  3)  which  is  not  on  the  curve ;  solve  the  two  equations  as 
simultaneous  and  thus  obtain  the  point  of  tangency  of  a  tan- 
gent from  (5,  3)  to  the  given  curve. 

5.  Find  tangent  and  normal  to  the  curve  x2— Wx— 8y— 5  =  0, 
at  the  point  whose  abscissa  is  2 ;  find  the  tangent  of  slope  —  2 
to  this  curve. 

6.  What    tangential    property  of   parabolic  curves   makes 
them  useful  in  reflectors  ?     Explain.     Prove  the  property. 

7.  Write  the  equation  of  a  hyperbola  having  the  foci  and 

x2      yz 
vertices  of  the  ellipse  —  4-  ^-  =  1  as  vertices  and  foci,  respec- 

—  '  *  . ' 

tively ;  find  where  these  curves  intersect ;  write  the  equation 
of  a  tangent  to  each  of  the  curves  at  one  of  the  points  of  inter- 
section ;  discuss  these  lines. 


TANGENTS  AND  NORMALS        345 

8.  Write  the  equation  of  the  tangent  at  a  point  (xl}  y^)  to 
each  of  the  following  curves  ;  use  the  general  formula  ;  time 
yourself. 

a.  4z2-6a;  +  9?/2  +  5?/  =  0. 

b.  4z2-6a;-9y2-5?/  =  0. 

c.  4  x2  —  6  x  —  5  y  =  0. 
d. 


cc        y 
e.    --  h  —  =  1  5  it  is  not  necessary  to  clear  of  fractions  as 

Y^g-  and  2*3-  can  be  thought*  of  as  co-efficients  of  x2  and  y2. 
/.    X2  _  QX  -4y2-  8^  +  7  =  0. 

9.   Find  the  tangents  to  the  first  three  curves  in  the  pre- 
ceding  exercise   at   the   points   where   these   curves  cut   the 


10.    Find   one  point  on   each  of   the  curves  of   the  eighth 
problem  and  write  the  equation  of  the  tangent  at  that  point. 


From  Tyrrell,  History  of  Bridge  Engineering 
Elliptical  arch  bridge  at  Hyde  Park  on  the  Hudson 

The  span  is  75  feet  and  the  rise  is  14.7  feet.     Note  that  the  reflection 
completes  the  ellipse. 


CHAPTER   XXII 

APPLICATIONS   OF  CONIC   SECTIONS 

1.  General.  —  Numerous  applications  of  the  conic  sections, 
viz.,  circle,  ellipse,  parabola,  and  hyperbola,  have  been  indi- 
cated in  the  problems  given  under  the  discussion  of  each  curve. 
In  general  it  is  the  tangential  properties  of  the  curves  and 
the  further  geometrical  peculiarities  of  these  curves  that  make 
them  so  widely  and  so  variously  useful.     The  fact  that  simple 
geometrical   properties   are    connected  with  curves  given  by 
algebraic   equations   of   the   first  and  second  degrees  in  two 
variables  seems  to  imply  a  certain  harmony  in  the  universe  of 
algebra  and  geometry. 

2.  Laws  of  the  universe.  —  In  1529  the  Polish  astronomer- 
mathematician,    Copernicus    (1473-1543),    rediscovered    and 
restated  the  fact,  known  to  ancient  Greeks,  that  the  sun   is 
the  center  of  the  universe  in  which  we  live ;  he  conceived  the 

346 


APPLICATIONS  OF  CONIC  SECTIONS  347 

planets  to  move  about  the  sun  in  circular  orbits.  About  a 
century  later  the  great  German  astronomer,  Kepler  (1571- 
1630),  was  able  to  establish  the  following  laws  of  the  universe : 

1.  The  orbits  of  the  planets  are  ellipses  with  the  sun  at  one 
focus. 

2.  Equal  areas  are  swept  out  in  equal  times,  by  radii  from 
the  sun  to  the  moving  planet. 

3.  The  square  of  the  time  of  revolution  of  any  planet  is 
proportional  to  the  cube  of  its  mean  distance  from  the  sun ; 

ip  2         ,-73 

i.e.  —  =  — ,  if  T\  and  T2  are  the  periodic  times  of  two  planets, 
7?      df 

and  dt  and  d2  the  diameters  of  their  respective  orbits. 

Kepler's  work  was  made  possible  by  that  of  all  his  pred- 
ecessors, particularly  the  Greek  mathematicians  who  had  so 
thoroughly  discussed  the  properties  of  the  conic  sections,  and 
further  by  the  work  of  the  Dane,  Tycho  Brahe  (1546-1601), 
whose  refined  observations  gave  the  necessary  data. 

Newton  (1642-1727)  completed  the  work  of  systematizing 
the  laws  of  motion  in  the  universe  in  which  we  live,  showing 
that  the  attraction  of  any  two  bodies  for  each  other  is  in- 
versely proportional  to  the  square  of  their  distance  apart  and 
directly  proportional  to  their  masses.  Newton  showed  further 
that  this  assumption  leads  to  the  elliptical  motion  in  the  case 
of  the  sun  and  any  planet. 

The  paths  of  comets  which  pass  but  once  are  known  to  be 
parabolas,  or  possibly  hyperbolas  with  eccentricity  close  to  1. 

3.  Projectiles.  —  The  first  approximation  to  the  path  of  a 
projectile  is  a  parabola.  Indeed  for  low  velocities,  below 
1000  feet  per  second,  the  path  is  almost  parabolic  even  with  air 
resistance.  The  parametric  equations  of  the  path  of  a  projec- 
tile shot  horizontally  with  a  velocity  of  1000  feet  per  second, 
neglecting  air  resistance,  are,  in  terms  of  t,  the  number  of 
seconds  of  flight,  as  follows  : 

x  =  1000 1, 

y  =  -  16 1\ 


348 


r  N I KIED   MATHEMATICS 


When  a  projectile  is  shot  at  an  angle  a  with  the  horizonal, 
we  have  shown  that  there  is  a  horizontal  component  of  veloc- 
ity,  v  cos  a,   and 
a  vertical  compo- 
nent of  velocity, 


tions  of  the  path 
of  this  projectile 
shot  from  the 
ground  as  avaxis 
are  as  follows : 

/*»  —  <i«     POS  f£  *  £ 

y  =  t*0  sin  a  •  t 

-  16  ft 

PROBLEM.  — 
Find  the  path  of  a 
projectile  thrown 
with  a  velocity 
of  50  feet  per 
second  horizon- 
tally from  the 
top  of  a  tower 
1000  feet  high. 


constitute  the  parametric  equations  of  the  path,  the  axes  being 
taken  through  the  top  of  the  tower.  Giving  to  t  values 
t  =  1,  2,  3,  •••  8,  these  equations  determine  the  position  of  the 
projectile  after  t  seconds. 

t  =  0,  1,  2,  3,  4,  5,  6,  7,  8  determines  the  following  points 
upon  the  parabola : 

(0,  0)  (50,  -16)  (100,  -64)  (150,  -144)  (200,  -256) 
(250,  -  400)  (300,  -  576)  (350,  -  784)  and  (400,  -  1024). 


Path  of  projectile  shot  horizontally  from  a  tower 

1000  feet  high;  initial  velocity  of  50  feet  per 

second 

X  =  50 1, 


APPLICATIONS  OF  CONIC  SECTIONS 


349 


Since  x  =  50  1,  y  =  -  16  £2,  y  =  -  16  •  (  —  Y,  for  all  values  of  t. 

\50J 

x2  =  —  156.25  y  is  the  equation  of  the  parabola  in  standard 
form  ;  the  coordinates  fa,  y^)  of  any  point  obtained  by  sub- 
stituting a  given  value  for  t  in  the  parametric  equations  above 

(x  \2  y 

—  •  }  =  —  77;. 
50J  16 

On  any  ordinary  coordinate  paper  the  curve  x2  =  —  156.25  y 
can  be  plotted  only  as  xz  =  —  156  y. 

The  drawing  shows  very  plainly  that  the  projectile  reaches 
the  earth  when  t  =  7.9  seconds,  approximately  ;  solving 

—  1000  =  -  16*2    (y  =  -16t2,  y  =  -1000), 
gives  t2  =  62.5, 

*  =  7.91- 

The  motion  of  a  falling  body  is  a  special  case  of  the  equa- 
tions above,  y  =  —  16  tz  gives  the  space  in  feet  covered  in  time 
t  seconds  by  a  freely  falling  body,  falling  from  rest. 

4.    Illustrative  problem.  —  For  a  bullet  shot  at  an  angle  of 
30°  with  a  velocity  of  1000  feet  per  second  the  equations  are  : 
x  =  866  t, 


This  bullet  will,  on  a  level  plain,  remain  in  the  air  until  y  =  0  ;  solving 
gives  the  value  for  the  time  of  flight.  The  range  is  given  by  inserting  the 
value  of  t  so  found  to  find  x. 

The  velocity  of  1000  feet  is  equiva- 
lent to  two  separate  velocities,  one 
vertical  of  500  feet  per  second,  one 
horizontal  of  866  feet  per  second. 
These  are  x  and  y  components  of  the 
velocity.  If  no  other  force  acted  on 
this  projectile,  the  path  would  be  a 

straight  line,  given  by 

Vertical  and  horizontal   compo- 

x  =  866  £,  nents  of  a  given  velocity 

y  =  500  t. 

But  since  gravity  acts,  diminishing  the  vertical  velocity,  the  total  y  is 
y  =  5W)t-16P, 


350  UNIFIED  MATHEMATICS 

the  —  16£2  being  due  to  the  effect  of  gravity.    The  fall  in  1  second  due 
to  gravity  is  independent  of  the  upward  motion. 
The  path  is  given  by 

x  =  866  «, 

y  =  500  t  -  16  V. 

When  y  —  0,  the  projectile  Is  on  the  ground,  the  se-axis. 

t\r\r\ 

«(500  -  16  0  =    0,   t  -  0,    or  t  =  —  ;  the  first  value,    t  =  0,  means 

16 

500 
simply  that  the  projectile  is  shot  from  the  ground,     t  =  —  =  31  £,  is  the 

number  of  seconds  the  projectile  is  in  the  air.  Finding  x  when  t  =  31J 
gives  the  horizontal  distance,  or  the  range.  Eliminating  t  gives  the 
Cartesian  form  of  the  equation  of  the  parabola 

x  =  866  t,  or  t  =  -^-,  whence 
866 

x  I  x  \* 

y  =  500 16 1 I  ,  which  reduces  to 

866          \SWj 

(x  -  13530)2  __  46870O  -  S906). 

The  numerical  work  is  somewhat  tedious  in  such  a  problem,  and  it 
is  indeed  in  most  practical  problems.  The  labor  can  be  materially 
shortened  by  remembering  that  since  the  initial  velocity  is  probably' 
correct  only  to  the  second  significant  figure,  correct  here  to  hundreds 
of  feet,  and  since  y  is  taken  as  32,  instead  of  32.2,  an  error  of  defect  in 
the  division  of  f  of  1  %,  the  error  by  excess  in  the  quotient  will  be  also 
f  of  1  %. 

PROBLEMS 

1.  Of  the  planets  Mercury  is  nearest  to  the  sun.     The  mean 
distance    of    Mercury    (=  a)    is  36    million   miles;  e  =  .2056; 
compute  the  equation  of  the  orbit  referred  to  the   principal 
diameter  as  axis ;  find  the  distance  of  the  sun  from  the  center 
of  the  path. 

2.  Venus  is  the  planet  which  is  second  in  order  of  distance 
from  the  sun ;  the  mean  distance  is  67.27  million  miles ;  e  = 
.0068,  compute  the  equation  and  constants  as  in  the  preceding 
problem. 

3.  Compute  the  orbit  of  Mars   and   focal   distance;  mean 
distance  is  141.7  million  miles  ;  e  =  .0933. 


APPLICATIONS  OF  CONIC  SECTIONS  351 

4.  Knowing  that  the  earth   has  a  time  of   revolution   of 
365.256  (use  365.3)  days  and  that  its  mean  distance  is  92.9 
million   miles,  compute  by  Kepler's  third  law  the  times  of 
revolution  of  the  planets  in  the  three  preceding  problems. 

5.  Discuss  the  maximum  and  minimum  speed  of  the  earth, 
assuming  that  the  angular  velocity  is  constant ;  note  that  the 
focal   distance  is  1.5   million   miles,   and  the  variation  will 
depend  upon  the  different  lengths  of  the  radius. 

6.  Assuming  that  the  big  gun  which  bombarded  Paris  had  a 
range  of  75  miles  when  pointed  at  an  angle  of  45°,  find  the 
initial  velocity  from  the  equations, 

x  =  .707  vt, 

y  =  .707  vt  —  16.1  V. 

Insert  in  these  equations  x  =  75  times  5280  and  y  =  0  and 
solve  for  v  and  t ;  the  values  obtained  are  the  theoretical 
initial  velocity  in  feet  per  second  and  the  time  of  flight  in  sec- 
onds, neglecting  air  resistance. 

7.  A  body  falls  a  distance  of  10,000  feet ;  find  the  time  of 
fall. 

8.  A  body  is  thrown  up  vertically  with  a  velocity  of  100 
feet  per  second  ;  discuss  the  motion. 

5.  Reflectors.  —  The  fact  that  the  tangent  to  a  parabola 
bisects  the  angle  between  a  focal  radius  and  a  line  parallel  to 
the  axis  leads  to  diverse  uses  of  the  parabola.  Rays  of  light 
from  the  sun  or  from  a  star  meet  a  parabolic  mirrored  surface 
whose  axis  is  directed  towards  the  sun  or  star  in  rays  parallel 
to  the  axis  of  the  parabolic  surface ;  these  rays  converge  at  the 
focus  of  the  parabola  and  are  by  this  means  intensified. 

In  an  automobile  reflector  and  in  searchlights  the  conditions 
are  reversed  ;  rays  emanating  from  the  central' light  at  the 
focus  are  reflected  in  rays  parallel  to  the  axis. 

A  ray  of  light  directed  towards  one  focus  of  a  hyperbolic 
surface  striking  the  surface  is  reflected  towards  the  other 


352 


UNIFIED  MATHEMATICS 


focus,  since  the  tangent  bisects  the  angle  between  the  focal 
radii.  This  property  is  used  by  astronomers  to  re-focus  the 
rays  of  light  from  the  parabolic  mirror  at  a  point  which  does 
not  lie  between  the  parabolic  mirror  and  the  sun  or  star.  An 
elliptical  mirror  beyond  Fl  might  be  used  for  the  same  purpose. 


Parabolic  and  hyperbolic  reflectors  at  the  Detroit  Observatory 
The  curvature  of  the  large  parabolic  mirror  is  greatly  exaggerated  on 

the  diagram. 

The  parabolic  mirror  here  pictured,  is  in  use  at  the  Detroit 
Observatory,  Ann  Arbor;  the  diameter  of  the  mirror  is  37.5 
inches ;  the  focal  length  is  19  feet ;  the  focal  length  of  the 
hyperbolic  mirror  used  is  5  feet ;  the  second  focus  of  the  hy- 
perbola is  2  feet  behind  the  vertex  of  the  parabola  and  at  this 
point,  Fz,  the  rays  are  directed  into  a  spectroscope. 

Sound  rays  are  entirely  similar  to  light  rays  so  far  as  reflect- 
ing properties  are  concerned.  In  an  auditorium  it  is  desired 
that  the  sound  waves  should  be  thrown  out  from  the  reflecting 
walls  about  the  stage  in  parallel  lines  to  all  parts  of  the  build- 
ing; the  reflecting  surfaces  have  parabolic  sections  with  the 
focus  at  the  center  of  the  stage. 

This  is  the  case  in  the  Hill  Auditorium  at  Ann  Arbor, 
Michigan ;  axial  sections  of  the  hall  made  by  planes  are  para- 
bolic in  form,  having  the  focus  at  the  center  of  the  stage. 


APPLICATIONS  OF  CONIC  SECTIONS 


353 


6.  Architectural  uses  of  conies.  —  The  intimate  connection 
between  beauty  of  form  and  numerical  relations  is  undoubtedly 
illustrated  by  the  "  golden  section."  The  most  satisfactory 
dimensions  of  a  rectangle  from  an  artistic  standpoint  are  such, 


Hell  Gate  bridge,  over  the  East  River,  New  York  City 

The  largest  parabolic  arch  in  the  world,  in  one  of  the  most  beautiful 
bridges  of  the  world  ;  the  arch  has  a  span  of  977.5  feet,  height  220  feet. 

so  it  is  accepted  by  those  qualified  to  judge,  that  the  longer 
dimension  is,  approximately,  to  the  shorter  as  the  shorter  is  to 
the  difference  between  the  two.  In  other  words,  if  the  width 
is  given,  the  desired  height  is  found  by  the  "  golden  section," 
i.e.  by  dividing  the  line  in  extreme  and  mean  ratio.  Thus 
for  width  40,  the  height  x  is  found  by  solving  the  equation, 

40          x 


x      40  —  x 

this  gives  a  quadratic  equation  for  x.  Note  that  if  a  square 
is  cut  off  at  one  end  of  this  rectangle  a  similar  rectangle 
remains ;  so  also  if  the  square  on  the  longer  side  is  added  to  the 
rectangle  a  larger  rectangle  similar  to  the  original  one  is 
formed. 


354 


UNIFIED  MATHEMATICS 


We  have  found  a  certain  connection  apparently  existing  be- 
tween simplicity  of  form  and  simplicity  of  algebraic  equation. 
Thus  the  straight  line  is  represented  by  the  simplest  algebraic 
equation  in  two  variables,  the  first  degree  equation ;  the  circle 
which  is  the  simplest  curved  line  to  construct  is  represented 
by  a  particularly  simple  type  of  quadratic  equation ;  to  other 


The  Williamsburg  bridge  over  the  East  River,  New  York 

Longest  suspension  bridge  in  the  world  ;  the  parabolic  arc  of  each 
18-inch  cable  is  1600  feet  in  span  by  180  feet  in  depth,  width  118  feet ; 
weight  of  the  whole  1600-foot  span  is  8000  tons.  Largest  traffic  of  any 
bridge  in  the  world. 


types  of  quadratic  equations  in  two  variables  correspond  only 
three  further  curves,  viz.,  ellipse,  parabola,  and  hyperbola. 
That  these  second-degree  curves,  the  conic  sections,  are 
artistically  satisfactory  is  evident  from  the  extended  use 
which  has  been  made  of  these  forms  by  artists,  ancient  and 
modern. 

In   the  construction   of  arches  it  is  found  that  beauty  of 
geometric   form   is  intimately  connected  with   simplicity  of 


APPLICATIONS  OF  CONIC  SECTIONS  355 

algebraic  equation.  The  parabola  and  the  ellipse  have  wide 
uses  in  construction  not  only  because  of  beauty  of  form,  but 
also  because  of  purely  mechanical  adaptation  to  the  stresses 
and  strains  caused  by  the  weight  of  arch  structures.  A 
recognized  authority l  on  bridge  building,  states  that  "  arches 
must  be  perfect  curves,"  and  warns  against  the  use  of  false 
ellipses. 

The  fact  that  in  many  of  the  greatest  bridges  of  the  world 
the  pure  ellipse  and  parabola  appear  so  frequently  is  an  indica- 
tion of  the  wide  acceptance  of  the  theory  that  elliptical  and 
parabolic  arches  are  beautiful  in  form.  The  great  Hell  Gate 
Bridge  of  New  York  has  for  the  main  arch  a  true  parabola 
(see  problem  11,  p.  317) ;  London  Bridge  has  five  elliptical 
arches  as  the  fundamental  part  of  the  sub-structure.  Even 
the  hyperbola  has  been  used,  but  that  only  rarely.  Let  it  be 
noted  that  partly  because  of  the  greater  ease  in  design  the 
circular  arch  is  much  more  common,  and  the  approximation 
to  ellipse  or  parabola  by  using  several  circular  arcs  with  dif- 
ferent centers  is  also  common. 

No  less  than  four  distinct  and  different  uses  of  the  parabolic 
arc  are  found  in  the  construction  of  bridges  and  trusses.  The 
suspension  bridge  with  a  parabolic  cable  is  one  type ;  the 
parabolic  arch  with  vertex  below  the  roadway  of  the  bridge 
is  a  second  use ;  the  parabolic  arch  intersecting  the  roadway 
is  the  third  type ;  and  the  parabolic  arch  entirely  above  the 
roadbed,  as  a  truss,  is  a  fourth  type. 

Elliptical  arches  and  less  frequently  parabolic  are  commonly 
used  in  the  design  of  large  foyers  of  theaters  and  in  other 
large  halls. 

Parabolic  and  pure  elliptic  arch  forms  are  used,  although 
not  as  frequently  as  circular  and  horseshoe  forms,  in  the 
design  of  sewers.  Even  complete  perfect  ellipses  have  been 
used  (see  problem  6,  p.  356). 

1  Mr.  G.  H.  Tyrrell,  of  Evanston,  Illinois,  Artistic  Bridge  Design, 
Chicago,  1912. 


356  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  Solve  the  quadratic  in  the  preceding  article  and  check 
on  the  diagram. 

2.  Continue  the  series  of  rectangles  to  the  right  and  find 
the  height  above  forty  which  satisfies  the  conditions  given  for 
beauty  of  form  of  the  rectangle. 

3.  The  Panther-Hollow  Bridge,  Pittsburg,  has  a  parabolic 
arch,  360  feet  in  span  with  a  rise  of  45  feet.     Draw  the  parab- 
ola.    Assuming   that   the   vertical   chords    are   spaced   every 
twenty  feet  and   rise   15   feet  above   the   vertex,  find   their 
lengths. 

4.  In  the  preceding  problem  the  smaller  arches  leading  to 
the  bridge  itself  are  probably  elliptical.     The  width  of  these 
arches  is  28  feet  and  the  height  of  the  arch  proper  about  8 
feet.     Draw  these  arches. 

5.  A  parabolic  sewer  arch  used  in  Harrisburg,  Pa.  (designed 
by  J.  H.  Fuertes)  has  dimensions  of   6  feet  in  width  by  4 
feet  high.     Construct  ten  points. 

6.  A  vertical  elliptical  sewer  in  Chicago,  Western  Avenue 
sewer,  constructed  1910,  has  dimensions  12  x  14  feet.     Draw 
the  figure. 

7.  Draw  an  elliptical  and  a  parabolic  arch,  width  100  feet, 
height  30  feet ;  compare. 

8.  Draw  to  scale  the  parabolic  arc  of  the  Williamsburg  sus- 
pension bridge,  1600  feet  in  span  by  180  feet  in  depth.     Find 
the  equation  in  simplest  form,  choosing  proper  axes.     Find 
the  lengths  of  four  vertical  chords  from  cable  to  the  tangent 
at  the  vertex  of  the  arc. 

7.  Elliptical  gears.  —  On  machines  such  as  shapers,  planers, 
punches,  and  the  like  the  actual  movement  during  the  opera- 
tion of  shaping,  planing,  or  punching  is  desired  to  be  slow  and 


APPLICATIONS  OF  CONIC  SECTIONS 


357 


steady,  and  the  return  motion  is  desired  to  be  much  more 
rapid.  Circular  gears  give  a  uniform  motion,  but  elliptical 
gears  permit  the  combination  of  slow  effective  movement  with 
quick  return.  The  two  ellipses  are  of  the  same  size  and  are 


Two  positions  of  elliptical  gears,  mounted  on  corresponding  foci 

Note  that  the  right-hand  ellipse  swings  through  a  large  angle,  in  this  po- 
sition, as  compared  with  the  left-hand  one. 

mounted  at  the  corresponding  foci.  In  every  position  then 
the  two  ellipses  will  be  in  contact  since  the  sum  of  the  focal 
distances  in  either  ellipse  equals  2  a,  always,  and  this  also 
equals  the  distance  between  the  two  fixed  foci  which  are 
on  the  axes  of  rotation. 

PROBLEMS 

1.  Draw  three  positions  of  two  elliptical  gears,  each  being 
an  ellipse  6  inches  by  10  inches.  Determine  maximum  and 
minimum  radii.  When  the  ellipse  is  turned  5  times  a  minute, 
what  is  the  fastest  linear  speed  of  a  point  on  either  ellipse  ? 
Note  that  it  is  given  by  using  as  radius  the  maximum  radius, 
FP  in  our  figure.  Find  the  slowest  speed. 


358 


UNIFIED  MATHEMATICS 


2.    In  the  Sandwich  hay-press  of  our  illustration  the  diame- 
ters of  the  ellipse  of  the  elliptical  gears  are  21T\  inches  and 


Elliptical  gears  on  a  hay  press;  the  slow  pressure  stroke 

ISy1^  inches  ;  plot  the  graph  and  discuss  maximum  and  mini- 
mum linear  speed,  given  that  the  angular  velocity  is  twenty  to 
twenty-two  revolutions  per  minute. 


Elliptical  gears  on  a  hay  press;  quick  return  motion 

8.  Applications  in  mechanics  and  physics.  —  The  applications 
of  the  conies  in  mechanics  and  physics  are  very  frequent. 
Thus  the  equation  giving  the  period  of  a  pendulum, 


t  = 


or  p  = 


7T2   .   I 


APPLICATIONS  OF  CONIC  SECTIONS  359 

(see  page  317)  is  the  equation  of  a  parabola,  when  g  is  taken  as 
constant.  Similarly  the  velocity  of  water  flowing  from  a  tube 
or  over  a  dam  depends  upon  the  height  or  head  of  water  above 
the  level  of  the  tube  or  dam  ;  the  relation  is  i>2  =  64.4  h,  where 
h  is  measured  in  feet  ;  this  also  is  a  parabolic  relation. 

The  bending-moment  at  any  given  section  of  a  beam  sup- 
ported at  both  ends  and  uniformly  loaded  varies  at  different 
points  on  the  beam,  being  greatest  at  the  middle.  These  mo- 
ments are  computed  graphically  in  the  case  of  a  bridge,  being 
given  by  a  so-called  parabola  of  moments.  This  parabola  for 
a  bridge  of  length  /,  uniformly  loaded  with  a  weight  of  w  per 
foot,  is  given  by  the  equation, 

M  =i  wP—  iw  •  x2, 

wherein  x  is  the  distance  from  the  center  of  the  bridge.  The 
parabola  is  plotted  across  the  length  I  of  the  bridge,  with  the 
vertical  ordinate  at  the  mid-point  representing  the  maximum 
moment. 

Thus  if  a  bridge  is  100  feet  wide  and  uniformly  loaded 
2  tons  per  foot,  the  moment  at  any  point  x  distance  from  the 
center  of  the  bridge  is  given  by  the  formula 

3f  =  2xl°7-  1000s2 


Draw  the  corresponding  parabola,  choosing  appropriate  units. 

When  a  rotating  wheel  is  stopped  by  the  application  of 
some  force  which  reduces  the  velocity  uniformly  per  second, 
the  equations  giving  the  number  of  revolutions  before  the 
wheel  comes  to  rest  correspond  closely  to  the  equations  of 
motion  of  a  body  moving  under  the  acceleration  of  gravity. 

$  =  w0t  -  1  Jet2, 

6  represents  numerically  the  angle  covered  in  time  t  seconds, 
the  body  having  an  initial  rotational  speed  of  w0  revolutions 
per  second  and  the  velocity  being  retarded  every  second  by 
k  revolutions  per  second.  Here  again  we  have  an  equation  be- 


360  UNIFIED  MATHEMATICS 

tween  0  and  t  represented  by  a  parabola.  The  time  in  which 
this  body  comes  to  rest  is  obtained  by  dividing  the  initial 
velocity  by  the  uniform  decrease  in  velocity  per  second,  i.e. 
by  the  acceleration  (or  retardation). 

The  relation  between  pressure  and  volume  of  a  perfect  gas, 
temperature  being  constant,  is  given  by  the  equation  : 

p  •  v  =  fc  ; 

in  words  the  volume  is  inversely  proportional  to  the  pressure. 
Plotting  points  gives  points  on  a  hyperbola  of  which  the  j>axis 
and  the  v-axis  are  the  asymptotes. 

Such  illustrations  could  be  multiplied,  but  many  relations 
of  this  character,  e.g.  the  ellipsoid  of  inertia,  require  consider- 
able technical  explanation  which  would  go  beyond  the  limits 
of  this  work. 

9.    Quadratic  function.  —  The  graph  of  the  quadratic  function, 
cue2  +  bx  -j-  c,  is  the  locus  of  the  equation, 
y  =  ax1  +  bx  -j-  c. 


11   \  • 

2a) 
2  -  4  ac 

4a 

a(x+  6Y. 

y  ^ 

(a 

4a 

b  V_ 

l^f  1_&2_4ac^ 

2a)  ~ 

i  y  T"     j      )' 

a\            4a     y 

The  graph  of  y  =  ax2  -f-  bx  -f-  c  is  a  parabola  ;  x  H  --  =  0  is 

2a 

the  axis.     If  a  is  positive  the  parabola  opens  up  ;  the  vertex 

is  VI  --  ,   --  -  i  ;  if  &2  —  4  ac  is  negative,  the  vertex  is 
\     2a  4«      J 

above  the  o;-axis  and  no  real  value  of  x  makes  y  =  0,  since  the 
graph  does  not  cut  the  axis.    If  a  is  negative,  the  parabola  opens 


APPLICATIONS  OF  CONIC  SECTIONS  361 

down ;  if  62  —  4  ac  is  negative,  the  vertex  is  below  the  or-axis 
and  again  the  graph  does  not  cut  the  o>-axis.  If  62  —  4  ac  =  0 
the  graph  is  tangent  to  the  #-axis.  Evidently  62  —  4  ac  <  0 
is  the  condition  that  ax2  +  bx  +  c  =  0  should  have  imaginary 
roots ;  62  —  4  ac  =  0  is  the  condition  for  equal  roots ;  and 
&2  —  4  ac  >  0  is  the  condition  for  real  roots. 

PROBLEMS 

Plot  to  the  same  axes  the  graphs  of  the  functions  in  prob- 
lems 1,  2,  and  3.     Discuss. 

1.  y  =  5x>+2x-  7. 

2.  y  =  5  x*  +  2  x  +  7. 

3.  y  =  5o?2+2aj  +  £. 

4.  If  a  wheel  is  rotating  at  the  rate  of  800  revolutions  per 
second,  and  a  force  acting  continuously  reduces  the  speed  each 
second  by  40  revolutions  per  second,  find  the  time  in  which  it 
will  stop,  and  the  number  of  revolutions  made  during  the 
retarded  motion.     Use  the  formula  given  in  articled. 

5.  Given  that  10  cubic  centimeters  of  air  are  subjected  to 
pressure,  at  a  pressure  of   1  atmosphere   the  volume   is   10, 
hence  pv  =  10  is  the  equation  connecting  volume  and  pressure. 
Plot  the  graph  of  this  for  values  of  p  from  ^  atmosphere  to  5 
atmospheres'  pressure. 

6.  Plot  the  parabola  of  moments  for  the  Hell  Gate  Bridge, 
assuming   a   uniform   loading  of  2  tons   per  foot.      Do  not 
reduce  tons  to  pounds,  but  use  ton-feet  as  units  of  moment. 
The  equation  is  M  =  4902  —  #2,  taking  980  as  the  length  of  the 
bridge. 

7.  Plot  the  parabola  of  moments  of   the   Panther-Hollow 
Bridge  in  problem  3  of   the  preceding  exercise,  assuming  2 
tons  per  foot  as  loading.     The  equation  is  M  =  1802  —  y?. 

8.  Find  the  equation  of  a  parabola  whose  focal  length  is 
19  feet.     Draw  the  graph  to  appropriate  scale.     This  is  the 
parabola  which,  revolved  about  its  axis,  gives  the  parabolic 


362  UNIFIED  MATHEMATICS 

reflector,  previously  mentioned,  which  is  in  use  at  the  Detroit 
Observatory.  (See  page  352.) 

Find  the  equation  of  a  hyperbola  which  has  the  same  focus 
as  this  parabola,  the  axis  of  the  parabola  as  transverse  axis,  and 
the  second  focus  on  the  axis  at  a  distance  of  2  feet  on  the  other 
side  of  the  vertex.  This  is  the  hyperbola  which,  revolved 
about  its  axis,  gives  the  hyperbolic  mirror  mentioned. 

The  parabolic  mirror  has  a  diameter  of  37  inches.     Find  the 

1  Q    K 

abscissa  for  the  ordinate  -  — ,  thus  finding  the  depth  of  the 

1_A 

mirror. 

9.  Plot  the  parabola  y1  =  70.02  x.  This  is  the  parabola 
which  is  fundamental  in  the  construction  of  the  Hill  Audito- 
rium. (See  page  352.)  The  plane  of  the  floor  cuts  the  side 
walls  in  this  curve  ;  so  also  the  intersection  of  the  ceiling  and  a 
plane  passed  vertically  through  the  main  aisle  of  the  hall. 
In  the  plans  the  computations  of  ordinates  for  given  abscissas 
are  made  to  the  thirty-second  of  an  inch.  Compute  the  focal 
ordinate.  This  is  the  radius  of  the  circular  arch  over  the 
stage.  Compute  the  ordinates  for  x  =  21,  26,  31,  51,  and  71, 
and  express  in  feet  and  inches. 

10.  The  Italian   amphitheaters   are,  in  general,  elliptical. 
The   Colosseum  in   Rome  (see  illustration,  page  288)   is  an 
ellipse  with  axes  of  615  and  510  feet.    Draw  the  graph  to  scale. 
On  the  same  diagram  and  with  the  same  center  and  axes  of 
reference  draw  the  arena,  of  which  the  dimensions  are  281 
feet  by  177  feet,  to  the  same  scale.     Note  that  the  minor  axis 
of  the  arena  is  almost  the  " golden  section"  of  the  major  axis, 
i.e.  177  is  approximately  a  mean  proportional   between  281 
and  281  less  177.     Find  the  mean  and  compare. 

11.  The  bridge  at  Hyde  Park  (see  illustration,  page  346)  is 
elliptical,  with  a  span  of  75  feet  and  an  arch  height  of  14.7 
feet.     Draw  this  elliptical  arch  to  scale. 


CHAPTER   XXIII 

POLES,  POLARS,  AND   DIAMETERS 

1.  Definition.  —  The  straight  line 

AXJ.X  +  Byiy  +  G(x  +  a?,)  +  F(y  +  yi)  +  C  =  0 
is  called  the  polar  of  P±  (xlf  y^  with  respect  to  the  conic 

Ax*  +  By2  +  2  Gx  +  2  Fy  +  C  =  0. 
The  point  Pv  (xl}  y^  is  called  the  pole  of  the  line. 

2.  Fundamental   property  of   polar   lines.  —  If   the   polar  of 
PV(XI,  2h)    with   respect   to   the   given  conic  passes   through 
P2(#2,  2/2)?  then  reciprocally  the  polar  of  P2(x2,  Vz)  will  pass 
through  Pfa,  3h). 

This  fundamental  property  of  polar  lines  enables  one  to 
prove  complicated  geometrical  theorems  for  conies  with  a 
minimum  of  machinery.  The  proof  of  the  theorem  is  itself 
simple,  for  substituting  in  the  polar  of -P^a?!,  y^,  the  co- 
ordinates (£2>  2/2))  we  have  that 

AXM  +  Byaa  +  G(x<t  +  arO  +  Ffa  +  yt)  +  C  =  0, 

if  the  polar  of  PI  passes  through  P2.  However  the  polar  of 
P2  is,  by  definition, 

Ax2x  +  By2y  +  Q(x  +  x2)  +  F(y  +  i/2)  +  C  =  0, 

and  substituting  (x1}  y^)  gives  precisely  the  preceding  expres- 
sion, which  is  of  value  0;  hence  PI(X},  y^)  is  on  the  polar  of 
-^2(^25  2/2)- 

3.  Geometric  properties  of  the  polar.  —  If  P1(o;1,  y^  lies  on  the 
curve,  the  polar  is  the  tangent  at  that  point.     (See  the  preced- 
ing chapter.) 

363 


364 


UNIFIED  MATHEMATICS 


If  PI  (x1}  3/1)  lies  outside  of  the  conic,  the  polar  is  the  chord 
of  contact  of  tangents  from  Pj. 

Let  P2(xz,  y2)  be  the  point  of  tangeucy  of  a  tangent  drawn 
from  PI  ; 

by  definition,  the  polar  of  P2  is  the  tangent  at  P2(#2, 2fe)  5 
by  construction,  the  polar  of  P2  passes  through  Pt ; 


The  polar  of  any  point  outside  a  conic  is  the  chord  of  contact 

by  the  fundamental  reciprocal  property,  since  the  polar  of  P2 
passes  through  P1?  the  polar  of  Pr  will  pass  through  P2. 

Similarly,-  calling  P3(ar3,  y3)  the  other  point  of  tangency,  the 
polar  of  P!  (x1}  y^)  will  pass  through  P3. 

Since  the  polar  of  Px  is  a  straight  line  and  since  it  passes 
through  the  two  points  of  tangency,  it  is  the  chord  of  con- 
tact joining  these  two  points. 

If  PI  (xi,  yi)  lies  inside,  or  outside,  or  on  the  conic,  the  polar 
is  the  locus  of  the  intersection  ~R(x',  y')  of  tangents  drawn  at 
the  extremities  of  any  chord  passing  through  P:. 

Let  R  (x',  yf)  be  the  intersection  of  tangents  at  the  extremities 
of  any  secant ; 


POLES,  POLARS,  AND  DIAMETERS  365 

then,  by  the  construction,  the  secant  drawn  is  the  chord  of 
contact  of  R  (xf,  yf)  ; 

by  construction,  the  polar  of  R  (xf,  y'}  passes  through  PI(XI}  y^  ; 

hence,  by  the  fundamental  reciprocal  property,  the  polar  of  P1 
will  pass  through  R(x',  y')  ; 

but  the  polar  of  Pv  is  a  straight  line  ; 

hence  the  locus  of  R  (xf,  yf)  is  a  straight  line,  the  polar  of 

PI  (xi,  y\\ 

By  pure  Euclidean  geometry  it  is  rather  complicated  to 
prove  that  the  locus  of  the  intersection  of  tangents  at  the  ex- 
tremities of  all  chords  of  a  circle  passing  through  a  fixed 
point  is  a  straight  line.  The  above  proves  this  property  for 
every  conic. 

4.  Diameter  :  definition  and  derivation.  —  The  locus  of  the 
midpoints  of  a  series  of  parallel  chords  in  any  conic  is  called 
a  diameter  of  the  conic. 

The  method,  applicable  to  any  equation  of  the  second  degree, 
is  given  for  a  special  case.  Find  the  diameter  bisecting  chords 
of  slope  3  in  the  ellipse 


g  ^  +  ^  ^  = 

Let  y  =  3  x  +  k 

represent  any  line  of  slope  3.     Solve,  as  simultaneous,  with 

9  x2  +  25  yz  =  900,  which  represents 
the  conic. 

Substituting, 

234  z2  +  150  kx  +  fc2  -  900  =  0 

is  an  equation  whose  roots  are  the  abscissas,  xt  and  a^,  of  the 
two  points  of  intersection. 

Solving,  _ 

_  _  150  ft  +  V(160  k)*  -  4(234)  (A?  -  900) 
Xl~  ~~  ~'a 


_  _  150  k  -  V(150  fc)2  -  4(234) (fe2  -  900) 
-'  ~468~ 


366  UNIFIED  MATHEMATICS 

For  the  midpoint  (#',  y')  of  the  chord, 


_ 


2  78 

the  midpoint  lies  on  the  chord 
y  =  3  x  +  k, 
hence,  „'  _  i  *'  _i_  i- 


The  middle  point  of  any  chord, 

y  =  3  x  +  fr, 
is  given  by 


'  =  +  — -,  and  these  two  equations 
26 


Any  diameter  of  a  parabola  is  parallel  to  the  axis  of  the  parabola 

constitute  parametric  equations  of  the  locus  of  the  midpoint. 
Eliminating  k  by  solving  for  k  and  substituting  (or  by  division 
here),  we  see  that  for  every  value  of  k  the  coordinates  of  the 
middle  point  satisfy  the  equation 

i1—- A* 


POLES,  POLARS,  AND  DIAMETERS 
Hence  the  middle  point  is  on  the  straight  line 


367 


By  precisely  similar  reasoning,  the  diameter  bisecting  chords 
of  slope  m  in  the  conic,  given  by 


is  Ax  +  mBy  +     G   +  mF  =  0. 

Applying  this  to  the  simplest  standard  forms  of  ellipse  and 
hyperbola,  we  see  that  every  diameter  of  a  central  conic  passes 
through  the  center  ;  applying  to  the  parabola,  y2  =  4  ax,  we 
obtain  my  —  2  a  =  0.  This  shows  that  the  diameter  of  any 
parabola  is  parallel  to  the  axis  of  the  parabola,  for  when  a 
parabola  is  rotated  or  moved  to  any  other  position  any  diame- 
ter moves  with  the  curve,  preserving  its  position  relative  to 
the  axis  of  the  curve. 

5.  Reciprocal  property  of  diameters  in  central  conies.  —  In  the 
conic  9  x2  -+-  25  y2  —  900  =  0  above,  the  diameter  bisecting 


Conjugate  diameters ;  each  bisects  all  chords  parallel  to  the  other 

chords   of   slope   3   has  the  slope   —  ^.    Now  the  diameter 
bisecting  chords  of  slope  —  ^  has  the  slope  3,  for  by  substi- 


368  UNIFIED  MATHEMATICS 

tution  in  the  general  formula,  we  have  9  x  —  y\(25)  y  =  0,  or 
y  =  3x. 

Each  of  these  diameters  bisects  chords  parallel  to  the  other. 
These  are  called  conjugate  diameters.  In  precisely  similar 
manner  this  property  can  be  established  in  any  ellipse  or 
hyperbola  for  the  diameter  bisecting  chords  of  slope  m.  Do 
this  for  the  diameter  of  the  general  conic  above. 

PROBLEMS 

1.  What  is  the  equation  of  the  chord  of  contact  (polar)  of 
(10,  0)  with  respect  to  9  x2  +  25  y2  =  225  ?     Solve  this  with  the 
curve.     This  gives  the  points  of  tangency  of  tangents  from 
(10,0).     Write  the  equation  of  the  tangent  at  each  of,  these 
points.    This  process  illustrates  an  analytic  method  for  finding 
a  tangent  from  an  external  point  to  any  conic, 

Ax*  +  By2  +  2  Gx  +  2  Fy  +  C  =  0.     Explain. 

2.  Find  the  equations  of  the  tangents  to  the  following  conies 
from  the  points  given ;  follow  the  method  of  problem  1 ;  time 
yourself. 

a.  */2-8x  =  0,  from  (-2,  6). 

&.  yz  _  8  x  -  6  y  -  10  =  0,  from  (-3,  5). 

c.  x*  +  yz  -  10  x  +  8  y  -  59  =  0,  from  (18,  6). 

d.  x*  -  4  y2  -  10  x  +  8  y  -  59  =  0,  from  (10,  5). 

e.  z2  +  y2  -  25  =  0,  from  (1,  8)  and  from  (2,  8). 

a/2     w2 

3.  Prove  in  the  ellipse   — f-  2-  =  1  that  if  the  diameter  is 

a2      62 

drawn  through  (xi}  ?/i),  the  tangents  at  the  extremities  of  this 
diameter  are  parallel  to  the  polar  of  PI(XI}  y^.  Call  the 
points  on  the  diameter  (x2,  y2)  and  (o^,  y3),  and  note  that  they 
lie  on  the  ellipse.  Write  the  equations  of  the  different  lines 
mentioned. 

4.  Prove  the  property  mentioned  in  the  preceding  problem 

a;2      v2 

for  the  hyperbola —  =  1  and  for  the  parabola  yz  —  4  ax  =  0. 

a2     &2 


lt 


POLES,  POLARS,  AND  DIAMETERS  369 

5.  In  problem  2,  write  the  equations  of  the  diameters  bisect- 
ing chords  of  slope  2   and   of   slope  —  |,    using  the  general 
formula  for  diameter. 

6.  In  problem  5  write  the  equations  of  the  conjugate  di- 
ameters in  the  central  conies. 

7.  Draw   the    circle    x?  +  y2  —  36  =  0  ;    draw    5    secants 
through  the  point  (4,  3)  ;  draw  the  tangents  at  the  two  inter- 
section points  of  each  secant  with  the  curve.     The  5  points 
of  intersection,  one  from  each  pair  of  tangents,  should  lie  in 
a  straight  line.     What  theorem  proves  this  ? 

8.  Prove  that  the  tangential  parallelogram  circumscribed 
at  the  ends  of  conjugate  diameters  of  an  ellipse  &2x2+a2t/2=a262 
has   a   constant   area.     First   show  that   the   one  end  of  the 
diameter    conjugate    to    the   diameter    through   PI(XI}  y^   is 

PJ  —  ^-,  —  -  )  ;  find  the  equation  of  the  tangent  at  P^  (x 

\       b      a  J 

find  the  distance  between  (0,  0)  and  (xl}  y^  ;  find  the  equation 
of  OPi  ;  find  the  perpendicular  distance  from  P2  to  OP\  ;  by 
multiplication  show  that  the  area  of  this  quarter  of  the  given 
parallelogram  is  constant. 

9.  Taking  Pi(xlt  y^  and  P^fa,  y2),  which,  by  the  preceding 
exercise,  may  be  written  PJ  —  ^±,  -^1  V.as  the  extremities  of 

a  pair  of  conjugate  diameters  in  the  ellipse  b2x2  +  azy2  =  a2&2, 
show  that  the  sum  of  the  squares  of  OP\  and  OPZ  equals 
a2  +  62. 

HINT.  —  Reduce  the  expressions  for  OPi2  and   OP22  to  common  de- 
nominator, and  use  the  fact  that  PI(XI,  y\}  is  on  the  given  ellipse. 

a;2      v2 

10.    In  the  hyperbola  --  —  =  1,  the  conjugate  diameter  to 
a2      62 

the  diameter  through  Pi(xt,  yj  on  the  hyperbola  does  not 
cut  the  curve  itself.  Prove  this. 

The   extremities   of   the   conjugate   diameter  are  taken  as 
the   points  in  which   the   conjugate   diameter   intersects   the 


370  UNIFIED  MATHEMATICS 


conjugate  hyperbola  -.  — «£«=— 1«      With  this  definition  the 
a2      62 

property  of  problem  8  can  be  proved  to  be  true  for  the  hyper- 
bola. State  the  method.  What  modification  would  you  expect 
so  far  as  the  property  of  problem  9  is  concerned  ? 

11.  Prove  that  the  polars  of  all  points  on  a  diameter  of  any 
conic  are  parallel,  comparing  with  problems  3  and  4  above. 

12.  Show  that  tangents  at  the  extremities  of  a  series  of 
parallel  chords  in  any  conic  intersect  on  the  corresponding 
diameter. 

13.  Prove   that  any   point   on   a   diameter   of   the   ellipse 

— h  —  =  1  and  the  intersection  of  the  polar  of  the  point  with 
a2  62 

the  diameter  divide  the  diameter  length  internally  and  ex- 
ternally in  the  same  ratio. 

HINT.  — Take  (zi,  j/i)  and  (—  Zi,  —  2/1)  on  the  curve  as  the  extremities 
of  the  diameter  ;  take  the  point  of  the  diameter  as  the  point  ( —  — -1 , 

y\  +  ryi\  whjch  divides  the  diameter  externally  in  the  ratio  r ;  find  the 
l-r  ) 

intersection  point  of  the  polar  of  this  point  with  y  =  —  z,  and  note  that  it 

is  the  same  as  the  point  which  divides  the  line  joining  (zi,  t/i)  to  (—  Z],  —  y\) 
internally  in  the  ratio  r.  The  property  holds  for  any  conic. 

If  through  any  point  a  secant  to  a  conic  is  drawn,  the  point 
and  the  intersection  of  the  polar  of  the  point  with  the  secant 
divide  the  chord  of  the  conic  formed  by  the  secant  internally 
and  externally  in  the  same  ratio.  The  proof  is  somewhat 
more  complicated  than  that  of  the  preceding  special  case. 

14.  Show  that  the  tangential  parallelogram  to  any  central 
conic  formed  by  the  tangents  at  the  extremities  of  a  pair  of 
conjugate  diameters  has  its  sides  bisected  by  the  points  of 
tangency.     An  ellipse  can  be  rather  neatly  inscribed  in  any 
parallelogram  by  drawing  the  ellipse  tangent  to  the  sides  of 
the  parallelogram  at  the  midpoints. 


CHAPTER   XXIV 

ALGEBRAIC  TRANSFORMATIONS  AND 
SUBSTITUTIONS 

1.  Transformation  of  coordinates.  —  For  varied  reasons  it  is 
sometimes  found  desirable  to  change  the  location  of  the 
coordinate  axes  with  respect  to  a  curve  which  is  given  by  an 
equation  involving  variables.  Usually  this  shifting  of  the  axes 
is  for  the  purpose  of  simplifying  the  discussion  of  the  geo- 
metrical properties  of  the  curve  in  question.  Thus  the  ellipse 

(x  _    fay  (y   _    fc\2 

has   been  given   in   the   form  ^  -  '-  +  >3.  —  —L.  =  1?   but  the 

geometrical  properties  of  the  same  curve  are  discussed  with 
reference   to  the  center  (h,  k)  as  origin,  giving  the  equation 


The  axes  may  be  subjected  to  a  translation,  giving  new  axes 
O'X'  and  0'  T'  parallel  to  the  old  axes  ;  or  the  axes  may  be 
turned  through  an  angle  a,  giving  new  axes  OX'  and  OY1 
about  the  old  origin  ;  the  two  motions  can  be  combined,  execut- 
ing first  the  translation,  usually,  and  then  the  rotation  ;  it  is 
possible  also  to  shift  to  new  axes  inclined  at  an  oblique  angle 
to  each  other,  but  the  formulas  involved  are  too  complicated 
for  an  elementary  work. 

2.  Translation  of  axes.  —  Suppose  the  a>-axis  fixed  and  the 
y-axis  moved  parallel  to  itself  to  a  new  origin  0'  at  distance 
00'  =  h,  from  0.  Take  P(x,  y)  as  the  coordinates  of  any 
point  with  reference  to  the  original  axes.  Evidently,  as  the 
a^axis  is  unchanged,  the  y  of  this  and  every  other  point  remains 

371 


372 


UNIFIED  MATHEMATICS 


the  same.  Let  M  be  the  foot  of  the  perpendicular  from  P  to 
the  avaxis  ;  then  by  our  fundamental  property  of  the  distances 
between  three  points  on  a  directed  line 

OM=  oa  +  ax. 

But     OM=x,     OO'  =  h, 

the  distance  either  posi- 
tive or  negative  00'; 
while  0'3f  =  xf  by  defini- 
tion. Hence,  whatever 
the  position  of  P(x,  y), 
Translation  of  axes  we  h&W, 

X  =  x'  +  h. 

Similarly,  if  the  avaxis  is  shifted   parallel  to  itself  by  an 
amount  k. 


The  two  equations 


x  =  x'  +  h, 


transform  any  equation   given  with  respect  to  any  axes,  to  a 
set  of  parallel  axes  having  the  point  (h,  k)  as  origin. 

3.    Algebraic  substitution  in  functions  of  one  variable. 

THEOREM.  —  Substitution  of  x'  -f-  h  for  x  in  any  algebraic  equa- 
tion of  type  a0xr-  +  fl^sc""1  -j-  •••  an_^x  +  an  =  0,  n  an  integer,  gives 
a  new  equation  whose  roots  are  h  less  than  the  roots  of  the  old. 

The  proof  of  this  theorem  depends  directly  upon  the  pre- 
ceding article.  The  substitution  x  =  x'  +  h  moves  the  y-axis 
h  units,  reducing  the  abscissas  of  all  points  by  h  if  h  is  positive 
and  increasing  them  by  —  h  if  h  is  negative. 

ILLUSTRATION.  —  If  the  graph  of  y  =  x3  —  2  x2  —  18  x  +  24  is  plotted, 
the  substitution  y  =  y  and  x  =  x'  +  4  simply  shifts  the  y-axis  4  units  to 
the  right,  thus  decreasing  the  numerical  value  of  each  root  by  4. 

The  new  equation  is 
y  =  (x1  +  4)3  _  2(x  +  4)2  -  18(x'  +  4)  +  24  =  x"  +  10  x'2  +  14  x'  -  16. 


TRANSFORMATIONS  AND  SUBSTITUTIONS      373 

Now  whatever  number  substituted  for  x  makes  x3  —  2  ar2  —  18  x  +  24  =  0, 
it  is  evident  that  4  less  substituted  for  x'  will  make 

(x'  +  4)3-2(x'  +4)2-  18(x'  +  4)+24  =  0. 

Graphically,  of  course,  as  we  have  indicated,  the  y-axis  has  been  pushed 
4  units  towards  the  right,  and  the  abscissa  of  each  point  of  intersection 
of  the  curve  with  the  x-axis  has  been  reduced  by  4. 


Graph  of  y  =  x3  -  2 x2  -  18x  +  24 

Similarly,  in  the  general  equation  above,  when  x'  -f-  h  is  sub- 
stituted for  x,  whatever  number  a  satisfies  the  original  equa- 
tion in  x,  a  —  h  will  satisfy  the  new  equation  in  x'. 

Substitution  of  x'  +  h  for  x  in  any  algebraic  equation  forms 
a  new  equation  in  x'  whose  roots  are  h  less  than  the  roots  of  the 
given  equation. 

This  type  of  substitution  is  used  to  facilitate  the  computa- 
tion of  roots  of  numerical  algebraic  equations. 

A  simple  method  of  constructing  the  new  equation  in  nu- 
merical equations  will  be  explained  below,  in  section  11  of  the 
next  chapter. 


374  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  Show  that  the  formulas  of  transformation  given  trans- 
form the  equations  of  ellipse  and  hyperbola  having  (h,  k)  as 
center  to  the  simpler  form  without  first-degree  terms. 

2.  Transform  the  equation  (y  —  3)2  =  S(x  +  2)  to  the  point 
(3,  —  2)  as  new  origin,  new  axes  parallel  to  the  old. 

3.  By  translation  of  axes  transform  the  equation 

x*-4xy  -6x  +  8y  —  10  =  0 
into  a  new  equation  in  which  the  first-degree  terms  are  lacking. 

4.  Find  the  equation  of  the  line  3y  —  4  a;  +  6  =0,  referred 
to  parallel  axes  through  the  point  (3,  2). 

5.  Compare  the  slope  of  a  straight  line  referred  to  new 
axes  by  translation,  with  the  slope  referred  to  the  old  axes. 
Compare  intercepts.     Compare  the  slope  of   a  tangent  at  a 
fixed  point  on  any  curve  with  respect  to  new  and  with  respect 
to  old  axes. 

6.  Given  the  expression  for  the  volume  of  1000  cu.  cm.  of 
mercury  at  0°  C.  when  heated  to  t°  C.,  v  =  1000  +  .0018  1  (see 
page  63),  transform   to  parallel   axes  with  the   point  (t  =  0, 
v  =  1000)  as  new  origin  ;    find  the  new   equation  in  v'  and  t. 
Does   v'  represent  volume  ? 

13  1 

7.  Given  v  =  1054  H  --  ,  the  velocity  in  feet  per  second  of 

\J& 

sound  in  air  at  t°  centigrade,  transform  to  parallel  axes  with 
(32°,  1054)  as  new  origin  ;  discuss  the  equation. 

8.  The   equation   y?  —  2  x*  -  18  x  +  24  =  0  (page  373)  was 
found  to  have  a  root  between  x  =  4  and  x  =  5  ;  transform 


to  parallel  axes  through  (0,  4)  and  the  new  equation  in  x'  will 
have  a  root  between  0  and  1.  Compute  this  root  to  tenths  by 
substitution. 

9.  The  equation  or*  -  2  xz  —  18  x  +  24  =  0  has  '  a  further 
root  between  1  and  2.  Compute  this  root  to  one  decimal  place 
by  the  process  explained  in  the  preceding  problem. 


TRANSFORMATIONS  AND  SUBSTITUTIONS      375 

10.  Find  the  roots  of  2  y?  +  6  x2  —  10  x  —  8  =  0,  as  in  prob- 
lems  8   and   9    by   considering   the   graph   of    the    equation 
y  =  2x3  +  6x2  —  10 #—  8,  when   referred   to   new  axes.     (See 
problem  4,  page  99.) 

11.  Transform   the   following   equations   to   parallel   axes, 
having  (h,  k)  as  the  new  origin ;  determine  (h,  Jc)  so  that  the 
terms  of  the  first  degree  shall  disappear. 

a.  5  x2  +  4  xy  —  f-  —  8  x  -  5  y  -  10  =  0. 

b.  5x*  +  4:xy  +  y2-8x-5y-W  =  Q. 

c.  xy  —  7  x  —  10  y  —  5  =  0. 

a.   4  xz  —  6  *  -  yz  -  8  y  —  10  =  0. 

12.  Transform    the    following    equations   to   parallel   axes 
having  (h,  Jc)  as  origin.     Can  you  determine  (h,  fc)  so  that  the 
terms  of  the  first  degree  shall  disappear  ?     Why  not  ?     (See 
problem  13.) 

a.   4  #2  _  e  x  _  g  y-  10  =  0. 


13.  Show  that  if  an  equation  of  the  second  degree  contains 
no  first-degree  terms,  the  origin  is  the  center  of  the  curve  by 
showing  that  if  (ajb  yx)  is  any  point  on  the  curve  (— xi}  —  y$ 
is  also  on  the  curve. 

4.  Rotation  of  axes.  —  The  formulas  for  sin  (a  -f  /?)  and 
cos  (a  +  ft)  give  very  neatly  the  relations  which  exist  between 
the  coordinates  (x,  y)  of  a  point  referred  to  the  old  axes  and 
the  coordinates  (xf,  y')  referred  to  the  new  axes.  Take  P(x,  y) 
any  point  referred  to  the  original  axes ;  let  a  be  the  angle  of 
rotation  through  which  the  axes  are  turned  ;  let  ft  be  the  angle 
which  the  line  OP  makes  with  the  x'  or  new  a; -axis.  By 
section  4,  chapter  15,  for  all  values  of  a  and  /?, 

cos  (a  +  ft)  =  cos  a  cos  ft  —  sin  a  sin  ft, 
OP  cos  (a  +  /8)  =  OP  cos  a  cos  ft  —  OP  sin  a  sin  /3. 


376  UNIFIED  MATHEMATICS 

But      OP  cos  (a  +  /?)  =  z;     OP  cos  0  =  a/;     OP  sin  /?  =  ?/'; 

hence, 

x  =  x'  cos  a  —  y  sin  a. 

Further,  sin  (a  +  ft)  =  sin  a  cos  /?  +  cos  a  sin  /?. 
Multiplying  by  OP,  and  substituting, 

y  =  x'  sin  a  +  y'  cos  a. 

These  same  relations  might  also  have  been  obtained  by  pro- 
jection ;  they  hold  for  every  position  of  the  point  P. 

x  —  x'  cos  a  —  y'  sin  «, 
y  =  x'  sin  a  +  y'  cos  a, 

effects  the  rotation  through  the  angle  a,  and  refers  any  equation 
in  two  variables  to  new  axes  inclined  at  an  angle  «  to  the  old 
axes. 

5.  Every  second-degree  equation  in  two  variables  represents  a 
conic  section.  Proof  —  To  prove  this  theorem  we  need  only  to 
show  that  the  equation 

(1)  Ax*  +  2  Hxy  +  By*  +  2Gx  +  2Fy  +  C=0 
can,  by  rotation  of  axes,  be  transformed  to  an  equation  of  the 
(2)  Ax*  +  W  +  2  Gx  +  2  Py  +  C  =  0. 

Every  equation  of  this  latter  type  represents,  as  we  have 
shown,  either  circle,  ellipse,  parabola,  or  hyperbola  or  some 
limiting  form  of  one  of  these  curves. 

Substituting, 

x  =  x  cos  a  —  y  sm  «, 

y  =  x'  sin  a  +  y'  cos  a, 

the  equation  Ax2  +  2  Hxy  +  Ef  +  2  Gx  +  2  Fy  +  C  =  0 
becomes 


A(x'  cos  a  —  y'  sin  «)2-j-  2  #(#'  cos  a  —  y'  sin  «)(x'  sin  «+  y'  cos  «) 
+  B(x'  sin  a  +  y'  cos  a)2  +  2  £(#'  cos  a  —  y'  sin  a) 
-f  2  P(x'  sin  a  +  ?/'  cos  a)  +  (7  =  0. 


TRANSFORMATIONS  AND  SUBSTITUTIONS      377 

Collecting  terms,  we  have, 

(A  cos2  a  -f-  B  sin2  a  +  2  H  cos  a  sin  a)x'2 
-\-(A  sin2 «  +  B  cos2  a  —  2  #  cos  a  sin  a)y'2 
-f  [  —  2  A  cos  a  sin  a  -f-  2  H(cos2  a  —  sin2 «)  -j-  2  5  cos  a  sin  a]x'yf 
+  (2  G  cos  a  +  2  Fsin  a.)^  +(2  ^cos  a  -  2  G  sin  a)/  +  C  =  0. 

Let    A'x'2  +  2  tf'zy  +  By2  +  2  6?  V  +  2  F'tf  +  C  =  0 
represent  this  equation. 

We  wish  to  show  that  it  is  always  possible  to  find  an  angle 
a  for  which  H'  becomes  0. 

2  If 

Setting    //' =  0,   leads    to   the   equation   tan2«=— , 

noting  that  cos2  a  —  sin2  a  =  cos  2  a  and  2  sin  a  cos  a  =  sin  2  a. 
Since  H,  A,  and  5  are  real  numbers  and  since  the  tangent  of 
an  angle  can  have  any  value  from  negative  to  positive  in- 
finity, it  follows  that  there  is  always  some  angle  2  a,  for  which 

2  ff  ' 

tan  2  a  =  —  .     There  are  in  fact  always  two  positive  angles, 

A  —  -D 

less  than  360°,  2  a  and  2  a  +  180°,  which  satisfy  the  given  re- 
lationship. By  turning  through  a,  or  a  -f  90°,  one  half  of 
either  of  these  angles,  the  equation  Ax2+2  Hxy-\-Byz-{-  •••  =  0, 
reduces  to  an  equation  of  the  type  A'x'2  +  B'y'*  +  •••  =  0, 
with  the  coefficient  of  the  x'y'  term  equal  to  0.  The  angle 
a  of  turning  can  always  be  selected  as  a  positive  acute 
angle,  since  if  tan  2  a  is  positive,  2  a  may  be  taken  as  an  acute 
angle,  and  if  tan  2  a  is  negative,  2  a  may  be  taken  as  an  obtuse 
angle  of  which  the  half-angle  a  will  be  acute. 

Illustrative  problem.  —  What  angle  of  rotation  will  remove 
the  xy  term  from  3  xz  -f-  6  xy  —  5  y-  =  100  ? 

tan2«  =  -^  =  «, 
A-  B     8 

cos  2  a  =  | ;  cos  a  =  V\  (1  +  cos  2  a)  = 


VlO 


sina  =  Vj(l  —  cos  2  a)  =  —  ^L- 
v'10 


378 


UNIFIED  MATHEMATICS 


We  select  a  acute,  as  noted,  hence  the  positive  values  of  the  radical  are 
taken.    The  formulas  of  transformation  become, 

x  =  -?-*'  -  -L^y1  =J-(3x'-  y'), 
VlO         VlO         VlO 


y  = 


Vio 


—  y'  =  -r  («'  +  3  j/')- 
VIo       Vio 


Substituting,  we  have, 


=  100. 


The  hyperbola 


=  100,  or  4  x'2  -  6  y  '2  =  100 


In  combining  terms,  do  not  write  the  expansion  but  preferably  combine 
like  terms  by  inspection. 

Here  the  coefficient  of  x'2  is  f  J  -f  |g  _  ^  ;  of  x'y'  the  coefficient  is 
~~  H  +  i§  ~~  l§»  or  0,  which  checks  ;  for  y'2  we  have  ^  —  f$  —  f§.  Our 
equation  becomes, 


25      16.67 

This  curve  is  plotted  with  reference  to  the  new  axes,  inclined  at  an  angle 
a,  tan  a  =  },  to  the  x-axis.  The  coordinates  of  a  point  (x',  y')  on  this 
curve,  considered  with  respect  to  the  new  axes,  satisfy  the  new  equation 


TRANSFORMATIONS  AND  SUBSTITUTIONS     379 

_^ —  =  1  ;  when  considered  with  reference  to  the  old  axes  as  (x,  y) , 

25     16.67 

the  coordinates  satisfy  the  original  equation.  Thus  the  coordinates  of  the 
intersection  with  the  original  x-axis  (5.8,  0)  satisfy  the  original  equation  ; 
this  point  with  reference  to  the  new  axes  has  the  coordinates 

5.8  x3 


x'  =  x  cos  a  +  y  sin  «  =  • 


VlO 
-5.8 


y1  =  —  x  sin  a  +  y  cos  a  = 

VlO 

or  (5.5,  —  1.8).  The  values  for  (x',  y')  in  terms  of  (x,  y)  can  be  con- 
ceived as  obtained  by  rotating  through  the  angle  —a. 

PROBLEMS 

1.  Find  the  equation  of  the  curve  xy  —  7x  +  3y—  15=0 
when  referred  to  axes  making  an  angle  of  45°  with  the  given 

axes.     Note    that    a    is    45° ;   sin  a  =  cos  a  =  — = ;   rationalize 

V2 

denominators  after  substituting.  Plot  the  new  axes  at  the 
angle  indicated  and  plot  the  graph  of  the  new  equation,  obtained 
by  substitution,  with  reference  to  the  new  axes. 

2.  Find  the  equation  of  the  curve 

9xz  +  24  xy  +  16  y2  -  6x  -  15  y  =  0 

with  reference  to  axes  making  an  angle  arctan  ^  with  the  old 
axes.  Note  that  sin  «  =  i  and  cos  «  =  f ;  in  substituting  take 
the  fraction  i  as  a  factor  in  the  value  of  both  x  and  y  and, 
after  substituting,  combine  terms  by  inspection  without  writ- 
ing each  expansion  separately. 

3.  Find  the  equation  of  the  curve 

59x2-2ixy  +  66  f-  +72a;-396?/  +  444  =0, 
when  referred  to  new  axes  such  that  the  new  avaxis  makes  an 
angle  whose  tangent  is  f  with  the  old  axis  of  abscissas. 

4.  In  the  equation  Axy  —  8x  +  10y  +  7  =  Q  make  the  gen- 
eral substitutions  which  effect  the  turning  of  the  axes  through 
an  angle  a,  and  determine  a  so  that  the  coefficient  of  the  x'y' 
term  shall  disappear. 


380  UNIFIED  MATHEMATICS 

6.    Nature  of  the  conic  Ax-  +2  Hxy  +  By*-  +  2  Gx  +  2  Fy  +  C=  0. 

A  central  conic  is  one  which  has  a  point  which  is  such  that 
every  chord  passing  through  this  point  is  bisected.  If  this 
point  be  taken  as  origin  of  coordinates,  it  follows  that  if  (#',  y') 
is  on  the  curve  (  —  x',  —  y')  is  also  on  the  curve.  A  substitu- 
tion, x  =  x'  +  h  and  y  =  y'  +  k,  which  causes  the  terms  of  the 
first  degree  in  our  equation  of  the  second  degree  to  disappear 
gives  the  equation  Ad"-  +  2  Hx'y'  +  By'2  +  C"  =  0.  Now  what- 
ever point  (x',  y')  satisfies  this  equation  (—  x',  —  y')  will  also 
satisfy  the  equation,  and  hence  the  new  origin  is  the  center  of 
this  conic. 

The  substitution  x  =  x'  +  h  and  y  =  y'-\-k  gives  two  linear 
expressions  in  h  and  k  as  coefficients  of  the  new  x'  term  and  y' 
term,  and  these  are  set  equal  to  zero  and  solved  for  h  and  k  to 

determine  the  center. 

/ 

2Ah+2Hk  +  2G=Q 
and  2Hh  +  2Bk  +  2F=0 


are  the  two  equations  which  determine  the  center. 

If  the  two  equations  which  serve  to  locate  the  center  repre- 
sent two  parallel  lines  in  h  and  k,  the  conic  has  no  center  and 

A  TT 

is  a  parabola.     This   condition   is   that  --  =  --  ,  or   that 

Hz  -  AB  =  0.     When  H*-AB=0,  the  terms  Ax2  +  2  Hxy  +  By* 
form  the  square  of  a  linear  expression  in  x  and  y. 

Further  it  is  shown  below  that  if  H2  —  AB  <  0,  the  conic  is 
an  ellipse,  and  if  H  2  —  AB  >  0,  the  conic  is  a  hyperbola.  The 
conditions  determining  the  nature  of  the  general  conic  are  as 

follows:  .  -T.      A     1V 

H-  —  AB  <  0,  ellipse, 

H2  —  AB  =  0,  parabola, 
H2  -  AB  >  0,  hyperbola. 

These  are  the  conditions  that  there  should  be  no  points  on 
the  curve  at  infinity,  one  point  at  infinity,  and  two  directions 
giving  infinite  points.  They  may  be  derived  by  substituting 


TRANSFORMATIONS  AND  SUBSTITUTIONS      381 

y  =  mx  +  Jc  and  determining  values  of  ra  for  which  the  quad- 
ratic has  infinite  roots  ;  it  follows  that  for  these  values  of  m 
the  line  y  =  mx  -f  k  will  meet  the  curve  in  points  infinitely 
distant.  For  the  ellipse  the  values  of  m  will  be  imaginary, 
and  H2  —  AB  <  0 ;  for  the  parabola  the  two  values  of  m  will 
coincide,  and  H2  —  AB  =  0  ;  for  the  hyperbola  the  two  values 
of  m  will  be  real  and  different,  representing  the  slopes  of  the 
two  asymptotes,  and  H2  —  AB  >  0. 

A  second  and  independent  proof  is  given  in  the  next  article. 
There  it  is  shown  that  the  product  A B'  is  positive  when 
H2  —  AB  is  negative ;  but  when  A'  and  B'  are  of  the  same 
sign  the  product  is  positive  and  the  curve  in  x'2  and  y'2,  not 
involving  x'y',  is  an  ellipse.  Similarly  the  product  A'B'  is 
negative  when  H2  —  AB  is  positive,  and  the  curve  represented 
by  the  transformed  equation  is  a  hyperbola. 

7.    Central  conies  ;  abbreviated  process  of  transformation.  — 

Substitution  method.  —  Determine  the  center ;  transform  to 
parallel  axes  with  the  center  as  new  origin ;  determine  a  and 
substitute  ;  plot  with  reference  to  the  final  axes. 

Abbreviated  method.  —  Determine  the  center  (h,  K) ;  trans- 
form to  (h,  K)  as  new  origin ;  determine  A'  and  B'  by  solving 
as  simultaneous  the  equations, 

A1  +  B'  =  A  +  B, 
-  A'B'  =  H2-AB; 

select  the  pair  of  values  of  A'  and  B'  such  that  A'—  B'  will  have 
the  same  sign  as  JET;  plot  the  new  equation  with  reference  to 
new  axes  having  the  origin  at  the  center  determined  and  the 
axes  inclined  at  an  angle  a  with  the  old  axes,  a  being  such  that 

tan  2  «  -      ~  ll 

~A^' 
Derivation  of  A'  +  B'  =  A  +  B ;  -  A'B'  =  H2  -  AB. 

* 

A'  =  A  cos2  a  +  B  sin2  a  +  2  H  cos  a  sin  a. 
B'  =  A  sin2  a  +  B  cos2  a—  2  H  cos  a  sin  a. 

By  addition,  A'  +  B'  =  A  +  B. 


382  UNIFIED  MATHEMATICS 

The  proof  that  —  A'B'  =  H*  —  AB  is  somewhat  long  but 
not  difficult.     To  the  product  —  A'B'  add 

H'2  =  [2  H  (cos2  a  -  sin2  a)  -  2  (A  -  B)  sin  a  cos  a]2, 

which  does  not  alter  the  value  since  H'  is  taken  to  equal  0. 
The  expressions  will  combine  to  H2  —  AB.  The  student  would 
do  well  to  verify  at  least  one  of  the  coefficients. 

Since  a  is  chosen  as  a  positive  acute  angle,  A'  —  B'  has  the 

same  sign  as  H,  for  A'  —  B' 

=  (A  -B)  cos  2  a  +  2  H  sin  2  a 

=2  H(A~  B cos 2  a  +  sin  2  A 
V   2H  J 

Now  sin  2  a  is  positive,  and   cos  2  a  has   the   same  sign  as 
and  hence  the  product  of  cos  2  a  by  — '-- —  is  positive  ; 


hence  A'  —  B'  is  the  product  of  2  H  by  the  sum  of  two  positive 
quantities  and  so  is  positive  if  H  is  positive  and  negative  if  H 
is  negative. 

The  equations  A'  +  B1  =  A  +  B 

-A'B'  =  H2  -  AB 

enable  us  to  determine  A'  and  B'  by  solving  these  as  simulta- 
neous equations.  Two  solutions  are  found,  and  the  solution 
is  selected  which  makes  A'  —  B'  have  the  same  sign  as  H. 

Only  the  new  constant  term,  when  transforming  to  (h,  k)  as 
new  origin,  offers  any  extended  computation.  This  constant 
term 

AW  +  2  Hhk  +  Bk*  +  2  Gh  +  2  Fk  +  C 

may  be  written 

h(Ah  +Hlc+G)+  Jc(Hh  +  Bk  +  F)  +  Gh  +  Fk  +  (7, 

which  reduces  to  Gh  +  Fk  -f  (7,  since  the  other  two  expressions 
within  parentheses  were  set  equal  to  zero  to  determine  the 
center. 

Illustrative  problem.  —  Find  center,  axes,  and  plot  the  conic, 
3  tf  +  6  xy  +  5  y2  -  12  x  —  18  y  -  24  =  0. 


TRANSFORMATIONS  AND  SUBSTITUTIONS      383 


Substituting  (x'  +  A,  y'  -f  k)  and  selecting  the  coefficients  of  x'  and  j/', 
to  set  equal  to  zero, 


6  h  +  10  k  -  18  =  0. 


Solving,  k  =  f ,  h  =  +  \. 


>x 


The  ellipse  3 x2  +  6  xy  +  5y2  -  12  x  -  18 y  -  24  =  0 

or  3  x'2  +  6  x'y'  +  5  y'2  -  —  =  0,  or  7.16  x"2  +  .84  y"2  =  40.5 
2 


C',  the  new  constant, 
new  (x',  y1)  equation  is 


=  -6.|-9.|-24=  —  ^.     The 


+  5j/'2-  —  =  0.     Note  that  tan  2  a  =—  =-3. 

A  —    £t 

A'+  B'  =  8, 
-  A'B'  =  9  -  15  =  -  6. 
Solving  by  substitution, 

-A'  (8  -A')  =-6, 
A'2  -  8  A'  +  6  =  0  ;  A'  =  4  ±  VlO. 
B'  =  4  T  VlO. 

J.'  —  B'  has  the  same  sign  as  H  ;  hence  the  upper  algebraic  signs  are 
taken,  A1  =  7.16,  B'  =  .84. 


384  UNIFIED  MATHEMATICS 

Our  final  equation  is 

7.16x"2+.842/"2 

•x"*      y"*  _  1 
5.66     48.3 

y"*     =1 


(2.38)2      (6.95)2 

Some  computation  is  unavoidable,  and,  in  general,  in  practical  applica- 
tions the  results  are  rarely  expressible  in  small  and  convenient  integers. 

PROBLEMS 

1.  Find  the  center,  axes,  and  plot  the  conic, 

5  a?  -  6  xy  +  3  if  +  12  x  -  6  y  —  30  =  0. 

2.  Plot  the  following  conies  by  turning  the  axes  through 

2  ff 
an  angle  a,  tan  2  a=  —   —  ,  so  as  to  eliminate  the  xy  term, 

and  thus  obtain  an  equation  to  plot  which  can  be  put  in  stand- 
ard form. 

a.  4  x2  +  4  xy  +  yz  —  6  x  +  8  y  —  12  =  0. 

b.  x2  —  4:xy  +  yz  +  2x  —  Wy  —  ll  =  0. 

c.  41  x2+  2±xy  +  3±y2-26x-32y-171  =  0. 

d.  4  xy  -  3  y*  -  7  x  —  10  y  -  15  =  0. 

3.  Apply  the  abbreviated  method  explained  in  section  6  to 
the  central  conies   in   the   preceding   problem;  compare   the 
numerical  work  involved  by  the  two  methods. 

4.  Find  five  points  on  the  first  and  second  conies  in  prob- 
lem 2  by  giving  values  to  x  and  computing  the  corresponding 
values  of  y. 

5.  Find  the  intercepts  with  the  coordinate  axes  of  each  of 
the  conies  in  problem  2  and  verify  your  graphical  construction 
by  these  points. 

6.  In  each  of  the  conies  of  problem  2  find  the  points  of 
intersection  with  the  line  y  =  mx  +  b  ;   determine  the  values 
of  ra  for  which  one  of  the  points  of  intersection  should  be  at 
an  infinite  distance.      In  the   case  of  the  hyperbolas    real 


TRANSFORMATIONS  AND  SUBSTITUTIONS      385 

values  of  m  will  be  found  ;  substitute  in  turn  each  of  these 
\values  for  m  and  determine  for  what  value  of  b  the  second 
point  of  intersection  will  move   off   to   an   infinite   distance. 
This  determines  the  two  asymptotes.     Explain. 

7.  Apply  the  abbreviated  method  to  the  discussion  of  the 
following  central  conies,  having  the  origin  as  center  : 

a.  x*  +  2  xy  +  4  yz  =  16. 

b.  4  a2  -6  ay  -3  ?/2  =  10. 

c.  2  a2  -4  a?/-?/2  =  -9. 

d.  5  z2  -  3  xy  +  ?/2  =  24. 

8.  In  the  hyperbolas  of  problem  2,  use  the  results  of  prob- 
lem 6  to  show  that  the  directions  of  the  asymptotes  are  given 
by  the  factors  of  the  terms  of  the.  second  degree. 

8.    The  hyperbola  as  related  to  its  asymptotes.  —  The  equation 
of  the  hyperbola  in  simplest  form, 

^_^_1 
a2     &2        ' 
may  also  be  written, 

(bx  —  ay)(bx  -f  ay)  =  a2&2, 
whence. 

bx  —  ay      bx  -f  ay         a262 

~~' 


Since  bx  —  ay  =  0  and  bx  -f  ay  =  0  represent  the  asymptotes 
of  this  hyperbola,  the  final  form  states  that  the  product  of  the 
perpendicular  distances  of  any  point  on  the  hyperbola  from 
the  two  asymptotes  is  constant.  The  converse  proposition  is 
also  true,  viz.,  if  a  point  moves  so  that  the  product  of  its 
distances  from  two  intersecting  lines  is  a  constant,  the  point 
moves  on  a  hyperbola  of  which  the  two  lines  are  the  asymp- 
totes. The  proof  of  the  converse  is  simply  that  the  bisectors 
of  the  angles  between  the  two  given  lines  could  be  selected  as 
axes  of  coordinates  and,  in  consequence,  the  two  lines  would 
have  as  equations,  expressions  of  the  form  y  —  mx  =  0  and 
y  +  mx.=  0.  Any  point  which  moves  so  that  the  product  of  its 


386  UNIFIED  MATHEMATICS 

distances  from  these  two  lines  is  a  constant  would  satisfy  the 
Cation  y-mx    .    y  +  mx  =  k . 

Vl  +  m2    Vl  +  m2 

but  this  equation  represents  a  hyperbola,  and   consequently 
the  given  locus  is  a  hyperbola. 

It  follows  from  the  above  argument  that  the  equation  of 
any  hyperbola  differs  by  a  constant  from  the  product  of  the 
first-degree  expressions  which,  put  equal  to  zero,  represent  its 
asymptotes.  The  terms  of  the  second  degree  in  the  hyperbola 
can  be  factored  always  into  real  linear  factors  in  x  and  y  (not 
necessarily  rational  so  far  as  the  coefficients  are  concerned) 
which  as  lines  have  the  slopes  of  the  asymptotes.  (See  problem 
8  of  the  preceding  list,  and  compare  article  6.)  A  particularly 
simple  type  of  hyperbola  equation  occurs  quite  frequently  in 
practical  problems  and  this  type  will  be  taken  to  illustrate  the 
method  which  is,  however,  general. 

Illustrative  example.  —  Plot  the  curve 

x 


y  = 


i-  x 

This  equation  may  be  written 


The  only  term  of  the  second  degree  is  xy.  Placing  the  factors  equal  to 
zero,  we  have  x  =  0  and  y  =  0.  The  asymptotes  are  parallel  to  our  co- 
ordinate axes.  The  equation  can  be  written  in  the  form 

(x  -h}(y  -k}  -  c. 
By  inspection  we  note  that  the  equation  may  be  written 

(y+l)(x-  !)  =  -!. 
The  asymptotes  are  given  by    y  +  1  =  0  and  x  —  1  =  0. 

The  intersection  point  is  the  center  of  the  given  curve  ;  further  points 
should  be  plotted  by  substitution  of  values  in  the  original  equation. 

This  equation  in  i  and  d,  i  =  -  ,  represents  the  relation  between  a 
1  —  d 

given  rate  of  discount  for  any  interval  and  the  corresponding  rate  of  in- 
terest. If  a  bank  in  lending  money  takes  out  interest  in  advance,  giving 


TRANSFORMATIONS  AND  SUBSTITUTIONS      387 

to  the  individual  not  the  face  of  the  loan  but  that  amount  less  the  interest 
upon  that  amount  for  the  given  interval  for  which  the  note  is  to  run,  the 
bank  is  said  to  discount,  the  note.  The  rate  of  interest  which  the  indi- 
vidual pays  is  obviously  greater  than  the  rate  d  which  is  used  as  the  dis- 
count rate  ;  the  relation  is 

«.JL. 

1  -d 

In  plotting  the  graph  of  this  curve  you  would  be  interested  only  in  values 
of  i  and  d  between  .01  and  .10,  and  you  would  confine  your  attention  to 
the  first  quadrant,  taking  i  inch  to  represent  .01  on  both  axes. 

PROBLEMS 

1.  Plot  the  curve  jp-/y  =  1000;  show  that  it   represents  a 
hyperbola  having  the  axes  as  asymptotes.     This  equation  rep- 
resents the  relation  between  the  pressure  and  volume  of  a 
quantity  of  gas  which  at  a  pressure  of  1  atmosphere  has  a 
volume   of   1000    cubic   units,    the    temperature    being   kept 
constant. 

2.  Discuss   the   nature    of    the    following   curves,   without 
making  any  transformation  of  axes ;  in  the   hyperbolas  give 
the  slopes  of  the  asymptotes,  and  in  the  parabolas  the  slope  of 
the  axis. 

a.   4  x2-  —  y*  —  8  y  =  0. 

6.  4  a2 -Sy- 10  =  0. 

c.  4  x2  -  4  xy  -  y2  - 100  =  0. 

d.  4:X2  —  4:Xy  +  yz  =  100  y. 

e.  4  a;2  —  4  a;?/ +  ?/2  =  100. 

f.  4  a2  -  4  a*/  +  4  y2  =  100. 

g.  4  x2  —  4  xy  —  10  x  =  25. 

h.  4ajy  — 7aj  +  10y  — 5  =  0. 

i.  xy  =  15. 

j,  4  a:2  +  4  y2  =  81. 

k.  3x2-12x-2y*-Wy-15  =  0. 

/.  3  a;2 -12  a;  +  2  ?/2  -  10 y- 15  =  0. 


388  UNIFIED  MATHEMATICS 

3.  Transform  to  new  axes  so  as  to  simplify  the  following 
equations  to  plot ;  select  the  appropriate  method   of   substi- 
tution adapted  to  each  equation. 

a.  a:2  +  12  xy  +  4  y*  —  4  x  —  24  y  —  10  =  0. 

b.  xi  +  3xy-3y2-10x-15y  +  2±  =  Q. 

c.  4  xz  —  4  xy  +  7  y2  —  10  y  +  4  x  —  25  =  0. 

d.  4  z2  -  12  a#  +  9  y2  —  6  x  -  10  =  0. 

e.  2xz  +  xy  +  2y2  -6x  +  6y  —  15  =  0. 
/.  a;2  +  4  *y  -  2  y2  -  8  x  +  20  y  —  30  =  0. 
gr.  a2  —  3  ay  —  7z  =  0. 

ft.   2  x2  -  6  a?  +  5  y2  -  20  y  -  10  =  0. 

4.  Given  that  an  aeroplane  covers  a  distance  of  one  hun- 

100 
dred  miles  in  t  hours,  its  velocity  "in  miles  per  hour  is  — — , 

t 

100 
i.e.  v  = ;   given  that  on  different  occasions  the  aeroplane 

L 

covers  100  miles  in  48  minutes  (.8  hours),  1  hour,  1  hour  6 
minutes,  l£  hours,  1  hour  and  24  minutes,  100  minutes,  and 
2  hours,  respectively,  find  the  velocities  and  plot  a  curve  giving 
the  relation  between  v  and  t.  Choose  units  so  that  you  can 
read  from  the  curve  between  the  extreme  values  the  velocity 
within  2  miles  per  hour  when  the  time  of  flight  for  100  miles 
is  given.  Note  that  the  curve  is  a  hyperbola. 

5.  Given  that  an  aeroplane  covers  on  one  trial  100  miles  in 
48  minutes,  on  another  trial  125  miles  in  61  minutes,  and  156 
miles  in  71  minutes  on  a  third  trial,  how  could  you  compare 
graphically  the  corresponding  velocities  ? 

6.  The  air  in  an  organ  pipe  vibrates   in  a  manner  some- 
what  similar  to   the  motion  of  a  pendulum ;  the  number  of 
such  vibrations  of  the  air   in   one  second   depends  upon  the 
length  of  the  pipe  and  upon  the  velocity  of  sound  in  air ;  the 

formula  7i  =  —  ,  v  in  feet  per  sec.  and  I  in  feet,  gives  quits 
- 1 


TRANSFORMATIONS  AND  SUBSTITUTIONS      389 


>  for 


closely  the  number  of  vibrations.     Plot  the  curve  n  = 

L  I 

values  of  I  from  1  to  20  feet,  choosing  appropriate  units.  The 
curve  gives  the  corresponding  number  of  vibrations  for  pipes 
of  different  lengths.  (See  section  3,  chapter  26.) 

7.  Discuss  fully  the  equation 

x*  -  2  xy  +  y*  —  10  y  =  0. 

8.  The  curve  of  transition  on  a  railroad  track  in  passing 
from  one  straight  track  to  another  is  sometimes  taken  as  para- 
bolic, because  of   the  fact  that  the  slope   changes  uniformly 


Parabolic  transition  curve  on  a  railroad  track 

The  parabolic  arc  is  used  for  vertical  as  well  as  .for  horizontal 

transition  curves. 

with  uniform  increases  of  the  horizontal  length  taken  parallel 
to  the  tangent  at  the  vertex  of  the  parabola.  Assuming  that 
the  track  AV  changes  its  direction  by  60°  to  VB  and  that  the 
transition  points  A  and  B  from  the  straight  line  to  the 
parabolas  are  taken  on  each  track  500  feet  from  the  point  of 
intersection  of  the  two  directions,  find  the  equation  of  the 
parabola.  Note  that  the  axis  FFis  inclined  at  an  angle  of 
120°  to  the  extension  of  AV;  note  that  E,  the  vertex  of  the 


390  UNIFIED  MATHEMATICS 

parabola,  is  midway  between  Vand  the  point  where  the  chord 
AB  cuts  the  axis,  since  the  tangent  to  a  parabola  cuts  off  from 
the  vertex  on  the  axis  a  distance  equal  to  the  distance  cut  off 
from  the  vertex  on  the  axis  by  the  perpendicular  to  the  axis 
from  the  point  of  taiigency.  Find  the  equation  of  the  curve 
with  respect  to  the  axis  of  the  parabola  as  ?/-axis  and  the  line 
through  V  at  an  angle  of  30°  with  AV  as  a?-axis ;  then 
transform  to  A  V  as  cc'-axis  and  a  perpendicular  to  AV  at  V 
as  y'-axis  by  turning  through  an  angle  of  —  30°,  using  the 
fundamental  formulas  for  rotation  of  axes. 

9.  Assuming  that  a  railroad  track  changes  its  direction  by 
40°,  30°,  20°,  and  10°  respectively,  find  the  equations  of  the 
parabolic  transition  curves  with  transition  points  (A  and  B,  as 
in  figure)  500  feet  from  the  intersection  point  of  the  two 
straight  tracks. 

10.  In  going   over  a  hill  the  form  of  curve   to  which  the 
track  bed  is  rounded  is  often  made  parabolic.     When  the  grade 
is  the  same  on  both  sides  of  the  highest  point,  the  problem  is 
precisely  that  of  finding  a  parabolic  arch.     Assuming  that  in 
a  horizontal  distance  of  5000  feet  the  hill  rises  100  feet,  find 
the  equation  of  the  parabola  having  the  vertex  at  the  highest 
point  and  passing  through  the  point  100  feet  lower  at  a  hori- 
zontal distance  of  5000  feet ;  find  the  four  intermediate  ordi- 
nates  at  distances  1000  feet  apart. 

11.  An   iron   wire   of   diameter  .2  cm.  and   length   I   cm., 
subjected  to  a  tension  T  caused  by  a  weight  W  grams,  when 
caused  to  vibrate  through  its  whole  length  has  the  number  of 
vibrations  determined  by  the  equation 


_1     / 

2i\ 


'.0777T 

When  the  weight  is  fixed  and  the  length  is  variable,  this  gives 
a  hyperbolic  relation  between  n  and  Z.     For  W  =  500  grams  the 

equation  is  approximately  n  = .     Plot  and  discuss. 


TRANSFORMATIONS  AND  SUBSTITUTIONS      391 

12.  In  the  preceding  problem  suppose  that  I  is  fixed  at  100 
centimeters  and  that  W  varies  between  100  grams  and  2000 
grams.      What  is  the  type  of  relationship  ?      What  would  be 
the  curve  obtained  by  plotting  to  iv-  and  w-axes  ? 

13.  The  deck  of  any  large  vessel  slopes  from  both  bow  and 
stern  downwards  towards  amidships.     The  vertical  section  of 
the  deck  from  bow  to  stern  consists.  of  two  parabolas,  having  a 
common  vertex  at  the  middle  of  the  ship.     Plot  the  parabolas 
which  are  used  for  a  vessel  400  feet  long,  having  the  highest 
point  at  the  bow  8  feet  above  the  vertex,  and  at  the  stern  the 
deck  4  feet  above  the  vertex.     Use  a  different  scale  for  y  than 
for  x,  —  at  least  twice  as  large. 

14.  Name  the  following  curves,  giving  such   facts  as  you 
can  by  inspection  : 

a.  3  x  +  2  y  -5  =  0.  ft.  (3a:+2y  —  5)(a?—  3)  =  10. 

b.  3x*  +  2  y-  5  =  0.  I.  3x*  +  3y2  =  Q. 

c.  3  x2  +  2  y1-  -  5  =  0.  m.  xy  -  7  x  +  6y  -  18  =  0> 

d.  3x*-5x  =  0. 


e.      x 

/.   3^4-3^  =  25.  p.      +    =  5. 

g.   3x*-6xy+3y*-5x=0.  y  r-= 


li. 

t  3*  +  2y  +  5,0.  r.   F=  331.7     l- 

j.    (3x+2y-5)(x-3)=0.  A/         273 

15.  The  highway  over  the  Michigan  Central  R.R..  tracks 
and  over  the  Huron  River,  on  the  Whitmore  Lake  road  near 
Ann  Arbor,  is  rounded  off  (in  profile)  to  a  parabolic  arc,  rising 
2.40  feet  in  a  span  of  240  feet.  Show  that  the  grade  leading 
up  to  the  arc  should  be  a  4  %  grade.  Draw  the  arc  to  scale. 


CHAPTER   XXV 

SOLUTION    OF   NUMERICAL   ALGEBRAIC    EQUATIONS 

1.  Solutions  of  algebraic  equations.  —  By  a  solution  of  an 
equation  of  the  type  a^x"  -J-  a1#n~1  +  a%xn~t  4.  ...  an_^c,  +  an  =  0, 
wherein  n  is  a  positive  integer  and  0$,  al}  a2  —  are  constants, 
we  understand  a  value  which,  substituted  for  x,  reduces  the  left- 
hand  member  to  zero.  That  such  a  solution  always  exists  is 
proved  by  methods  of  higher  mathematics.  The  theorem  that 
every  such  rational  integral  algebraic  equation  has  a  root  is 
called  the  fundamental  theorem  of  algebra;  it  was  first  proved 
about  a  century  ago  by  Gauss.  The  solution  may  be  a  real 
number  or  a  complex  number,  and  any  constant  coefficient  may 
be  real  or  complex ;  the  latter  involves  the  square  root  of  a 
negative  quantity  and  so  is  not  representable  as  the  abscissa 
of  any  point  on  our  axis  of  positive  and  negative  real 
numbers. 

Certain  types  of  algebraic  equations  are  solvable  in  terms 
of  the  general  constants  which  enter  as  coefficients.  Thus 
ax  +  b  =  0  is  solvable  in  terms  of  a  and  &,  and  ax*  -+-  bx  +  c  =  0 
is  solvable  in  terms  of  a,  b,  and  c.  It  has  been  shown  that  the 
general  cubic  in  one  variable  and  the  general  biquadratic,  or 
fourth  degree  equation,  are  solvable  in  this  way,  but  the 
general  equations  of  higher  degree  than  the  fourth  are  not 
solvable  in  this  sense. 

The  approximate  numerical  solution  of  the  real  roots  of 
rational  integral  equations  with  numerical  coefficients  is  readily 
obtained  and  we  have  indicated  in  Chapter  II  and  again  in  the 
preceding  chapter,  section  3,  problems  8-10,  the  general 
method  by  which  such  solutions  are  obtained  by  substitution. 

392 


NUMERICAL  ALGEBRAIC  EQUATIONS 


393 


Simplifications  for  purposes  of  computation  will  be  explained 
in  this  chapter. 

2.  Continuity.  —  The  height  of  an  individual  is  a  continuous 
function  of  the  age  of  the  individual ;  by  this  we  mean  that 
in  passing  from  one  height  to  another  the  individual  passes 
through  every  intermediate  height.  A  graph  representing  age 
as  abscissas  and  heights  as  ordinates  will  be  a  continuously 


Four  continuous  graphs.     One  discontinuous 
Continuity  in  passing  from  a  positive  to  a  negative  value. 

connected  curve.  Upon  this  curve  corresponding  to  any 
selected  age,  al}  a  period  of  time,  there  will  be  one  and  only 
one  corresponding  height,  A1}  and  corresponding  to  any  second 
age,  02,  a  second  ordinate,  7t2,  representing  height.  The  curve 
joining  the  two  points  (a},  Tij)  to  (a.2,  A2)  will  be  continuous  and 
every  intermediate  height  between  the  two  given  will  be 
found  to  be  represented  by  the  ordinate  corresponding  to  some 
age  intermediate  between  the  two  given  ages. 
The  rational  integral  function  of  x, 


in  which  n  is  any  positive  integer,  is  continuous  between 
any  two  values  of  x,  and  will  be  represented  by  a  continuous 
curve.  This  has  been  assumed  in  drawing  the  graph  of 
y  =  x3  —  2  x2  —  18  x  +  24,  and  in  other  graphs.  The  proof  in- 
volves discussion  somewhat  too  detailed  and  mathematically 
refined  for  an  elementary  course. 

The  symbol  f(x)  will  be  used  throughout  the  remainder  of 
this  chapter  to  represent  a  rational  integral  function  of  X  of 
the  type  mentioned  above. 


394  UNIFIED  MATHEMATICS 

3.  Graph  of  y  =f(x)  by  location  of  points. 

Give  to  x  the  appropriate  values,  find  the  corresponding  values 
of  y,  and  plot  the  points,  connecting  by  a  smooth  curve.  (See 
pages  70-71.) 

Apply  the  remainder  theorem,  and  employ  synthetic 
division  to  determine  values  of  the  function  corresponding  to 
given  values  of  x. 

4.  Remainder  theorem  and  synthetic  division.     (See  page  25.) 

When  j\x)  is  divided  by  x  —  a,  the  remainder  obtained  by  con- 
tinuing the  division  until  the  remainder  does  not  contain  x  is 
equal  to  the  original  expression  with  a  put  for  x. 

To  divide  f(x)  by  x  —  a,  employing  synthetic  division, 

a.  Arrange  f(x)  in  descending  powers  of  x  and  write  the 
coefficients  horizontally,  including  zero  coefficients  for  missing 
powers  below  the  highest  power  which  occurs. 

6.  Write  +  a  under  x  —  a,  the  divisor,  placed  at  the  left. 
Under  the  coefficients  of  f(x)  as  written  leave  space  for  a 
second  horizontal  row  and  draw  a  horizontal  line. 

c.  Under  the  coefficient  of  the  highest  power  of  x,  below 
the  horizontal  line  drawn,  place  this  coefficient  again.  Mul- 
tiply by  +  a  and  add  to  the  following  coefficient  to  the  right. 
Place  the  sum  below  the  line,  vertically  under  the  second 
coefficient;  use  this  number  below  the  line  as  multiplier  of 
+  a,  and  add  the  product  to  the  third  coefficient  and  continue 
this  process  until  you  have  placed  numbers  under  every  coeffi- 
cient (and  the  constant  term)  of  the  upper  row.  The  final 
number  which  appears  is  the  remainder  and  should  be  cut  off 
by  a  vertical  separator ;  the  numbers  under  the  horizontal  line 
are  coefficients  in  order  from  left  to  right  of  the  quotient 
when  f(x)  is  divided  by  x  —  a. 

Throughout  this  discussion  a  may  be  either  positive  or 
negative. 


NUMERICAL  ALGEBRAIC  EQUATIONS          395 

ILLUSTRATIVE  PROBLEM.  —  Divide  x3  —  2  x2  —  18  X  +  24  by  x  —  3,  and 
use  the  remainder  theorem  to  determine  the  value  of  this  function  of  x 
when  x  =  3.  x3  -  2  x2  -  18  x  +  24 

x  _  3)  1  -  2      -  18     +24 
+  3)     +3      +3-45 
1  +  1       -  15  (-  21 

x2  +  x  —  15  is  the  quotient  and  —  21  is  remainder;  —  21  is  the  value  of 
x3  —  2  x2  —  18  x  +  24  when  3  is  substituted  for  x.  Since 

xs  _  2  x2  -  18  x  +  24  =  (x2  +  x  -  15)  (x  -  3)  -  21, 
we  have,  substituting  3, 

33  _  2.  32- 18- 3  +  24=(32  +  3-  15)(3-3)-21 
=  0-21. 

PROBLEMS 

1.  Locate  ten  points  upon  the  graph  of  y  =2  a^-j-3  x2—  9  x— 7. 
Take  the  ten  points  between  x  —  —  4  and  x  =  +  4,  including 

\  and  —  \ ;  use  the  synthetic  division  method  of  finding  the 
value  of  y  except  for  x  =  0,  x  —  1,  and  x  —  —  1.  Plot  the 
points  and  draw  a  smooth  curve  connecting  them  ;  choose  the 
y  scale  so  as  to  keep  the  points  on  the  paper.  Locate  the  zeros 
of  the  function  on  the  graph. 

2.  Plot  the  graph  of  the  function  2  a?  +  3  x2  —  7  between  —  3 
and  +-3. 

3.  Plot   the   graph   of   the   function    2  y?  —  9  x  —  7.     Note 
where  the  graph  crosses  the  axis  of  x,  thus  locating  the  roots 
of   the   equation  2^  —  9^  —  7  =  0.       Factor   2  y?  —  9  x  —  7, 
dividing  by  the  factor  corresponding  to  the  rational  root  which 
you  have  found;  solve  the  resulting  quadratic,  and  compare 
with  the  values  found  by  the  graph. 

4.  Plot  the  graph  of  the  function  x4  -2  a?  +-3  z2-18  x+21 ; 
select  the  appropriate  interval  to  give  the  points  of  intersection 
with  the  ic-axis. 

5.  Plot  the  graph  of  y  =  x4  —  3  #2  —  21 ;  locate  the  zeros  of 
the  function  on  the  graph.     Solve  as  an  equation  in  quadratic 
form   x*  —  3  x2  —  21  =  0  and  compare  the  solutions   obtained 
with  the  roots  located  graphically. 


396  UNIFIED  MATHEMATICS 

6.  Plot  the  graph  of  the  function  4a^—  3ar+  .5  in  tin-  in- 
terval from  —  1  to  +  1 ;  substitute  for  x  the  values  —  1,  —  .8, 
-.5,  -.3,  —  .1,  0,  .1,  .2,  .3,  .4,  .5,  .6,  .7,  .8,  .9,  and  1,  finding 
the  values  in  general,  by  the  division  method  applying  the  re- 
mainder theorem.     The  roots  of  this  equation  represent  the 
values  of  sin  10°,  sin  50°,  and  sin  —  70°.     (See  section  10,  below.) 

7.  Locate  one  root  between  0  and  .1  of  the  equation 

4  s» -3  a;  4- .05234=0, 

by  substituting  for  x  the  values  0,  .01,  .02,  .03,  up  to  .1.  The 
value  .05234  is  the  sine  of  three  degrees  which  we  obtained  in 
problem  5,  page  245.  One  root  of  this  equation  gives  the  sine 
of  1°. 

5.  Number  of  roots.  —  A  value  of  ab  for  which  /(c^)  =  0,  is  a 
root  of  f(x)  =  0.      The  remainder  theorem  applies  and  conse- 
quently (x  —  aj)  is  a  factor  of  f(x)  since  the  remainder  when 
f(x)  is  divided  by  x  —  at  will  be  zero.     Nothing  in  our  argu- 
ment requires  that  al  be  a  real  number.    Hence,  dividing  f(x)  by 
(x  —  av),  a  new  equation  of  degree  one  less  will  be  obtained. 
This  equation,  by  the  fundamental  theorem  of  algebra,  also  has 
a  root,  02,  giving  a  quotient  of  degree  n  —  1.     The  number  of 
such  factors  corresponds  to  the  degree  of  the  equation,  n. 

Every  rational  integral  equation  of  the  nth  degree  has  n  roots, 
and  no  more.  For  no  further  value  of  x  could  make  the  product, 
a(x  —  a,i)(x  —  a*i)(x  —  a3)  •••  (x—  an),  equal  zero  without  making 
one  of  the  factors  zero  and  thus  coinciding  with  one  of  the  roots 
given. 

6.  Graphical  location  of  real  roots. — Any  real  root  of  a  rational 
integral  function  of  x  equated  to  zero  is  a  value  of  x  which 
makes  the  ordinate  in  y  =/(#)  equal  to  zero.     The  points  in 
which  the  graph  of  the  function  of  x  crosses,  or  touches,  the 
a;-axis  correspond  to  real  roots  of  the  equation,  f(x)  =  0,  or 
zeros  of  the  function. 

Our  assumption  of  continuity  enables  us  to  formulate  the 
following  theorem : 


NUMERICAL  ALGEBRAIC  EQUATIONS          397 


Betiveen  any  tico  values  x=  a  and  x  =  b,  for  which  the  two 
corresponding  values  of  f(x)  are  opposite  in  sign,  there  lies  at 
least  one  real  root  of  the  equation  f(x)  —  0. 


A: 


Fpur  graphs  passing  continuously  from  y  =  -  toi/=  —  ;  one  graph  with  a 

2  3 

discontinuity 

Thus  to  change  continuously  from  +  \  to  —  ^,  or  from  any  positive 
value  to  any  negative  value,  the  function  must  pass  through  all  values  in- 
termediate, including  0.  At  this  point  where  the  function  of  x  is  0,  the 
graph  of  y  =/(x)  crosses  the  axis. 

Illustrative  problem.  —  Locate  the  roots  of 


Plot  the  graph  of  y  =  x3  -  2  x2  —  18  x  -f  24  by  location  of  points.  Give 
to  x  values  from  —  5  to  +5,  find  the  corresponding  values  of  y,  and  plot 
the  points,  connecting  by  a  smooth  curve.  (See  page  71.)  Between  x=l 
and  x  =  2,  /(x)  changes  from  -f  5  to  —  12  ;  there  is  a  root  between  x  =  1 
and  x  =  2  ;  between  x  =  4  and  x  =  5  there  is  a  root,  as/  (4)  is  —  16  and 
/(5)  is  +  9  ;  at  x  =  —  4  there  is  a  root,  as/(—  4)  is  0. 

7.  Slope  of  y=f(x).  —  The  (h,  Jc)  method  of  finding  the 
tangent  at  a  point  (a^,  y^)  on  a  curve  applies,  as  we  have 
stated  in  Chapter  18,  section  11,  to  the  graph  of  a  rational  in- 
tegral function  of  x. 

Thus  in  y  =  x*  —  2xz  —  18  a;  +  24,  let  (x1}  y^)  be  any  point 
on  the  curve  and  (a^  +  /*,  yt  +  fc)  a  second  point.  It  is  desired 
to  find  the  slope  of  the  graph  at  (a;,,  y^ 


and 


y,  =  a^8  -  2a;!2  -  18a^  +  24, 
+  fc  =(a?i  +  h)3  -  2(xj_  +  h}2  -  l&fa  +  K)  +24,  since  (x1}  y,) 


398  UNIFIED  MATHEMATICS 

and  fa  +  h,  yl  +  K)  are  on  the  curve.     Subtracting  the  upper 
from  the  lower  equation,  member  for  member,  we  have, 

fc  =  7t(3  xf  -4^-18)  +  7i2(3  a,  -  2)  +  7i3, 


=  3  x?  -  4  xj.  -  18  +  h(3  xl  -  2)  +  &• 
h 

Let  h  approach  zero  ;  the  terms  on  the  right  containing  h  and 
h?  will  also  approach  zero,  as  the  coefficients  are  constants. 

Limit  -  =  3  x?  —  4  Xj.  -  18,  as  h  =  0. 
h 

"When  xl  =  1,  the  slope  of  the  curve  is  —  19  ;  when  xt  =  3, 
the  slope  is  —  3  ;  when  x±  =  4,  the  slope  is  +  14. 

A  double  root  of  any  equation  corresponds  to  a  point  at 
which  the  function  is  zero  and  the  slope  of  the  curve,  ob- 
tained by  the  (h,  k)  method,  is  zero. 

8.    Slope  and   maximum    and    minimum   points.  —  When   the 

slope  is  zero,  the  curve  is  for  the  instant  parallel  to  the  cc-axis. 
This  is  a  necessary  condition  for  a  maximum  or  minimum 
point,  i.e.  a  point  at  which  the  value  of  the  function  attains  a 
greatest  or  a  least  value  in  some  interval  which  includes  the 
point. 

This  may  be  accepted  by  the  student  as  graphically  evident. 
A  formal  proof  depends  on  the  methods  of  the  calculus,  and 
rests  essentially  on  the  method  used  in  finding  the  slope. 

PROBLEMS 

See  the  preceding  list  of  problems. 

1.  Find  the  slope  at  any  point  (x1}  y^)  of  each  of  the  follow- 
ing curves  and  locate  the  maximum  and  minimum  points 
on  the  curve  by  setting  the  slope  equal  to  0  and  solving  for  #x  : 


a.  y  = 

b.  y  =  2x*  +  3x*  —  7.  e.   y  =  4a»  —  3*  +  .5. 

c.  y  =  x*-3xz-21.  /.   y  =  4  x?  -  3  x  +  .05234. 


NUMERICAL  ALGEBRAIC  EQUATIONS          399 

2.   Find  the  slope  at  any  point  (xi}  y})  on 

y  =  x*  -  2  x3  +  3  x2  -  18  x  +  21. 

This  gives  the  slope  m  as  m  =  4  a?!3  —  6  x-p  -f  6  a^  —  18.  Plot 
the  graph  of  y  =  4  x3  —  6  x2  +  6  x  —  18,  and  note  that  the  zeros 
of  this  function  are  the  values  of  a^  for  which  the  slope  of  the 
curve  y  =  x*  —  2  y?  +  3  xz  —  18  x  +  21  is  0.  These  are  values 
of  #  for  which  the  original  function  has  maximum  and 
minimum  values. 

9.  Historical  note.  —  The  solution  early  in  the  sixteenth  cen- 
tury of  the  cubic  and  biquadratic  was  the  undisputed  achieve- 
ment of  a  group  of  Italian  mathematicians.     Fiori,  Tartaglia, 
and  Cardan  were  involved  in  the  solution  of  the  cubic,  while 
Ferrari,  pupil  of  Cardan,  solved  the  quartic.     Not  until  the 
beginning  of  the  nineteenth  century  was  it  shown  that  the 
general  equations  of  higher  degree  are  not  solvable,  this  being 
the  work  of  a  brilliant  young  Norwegian  named  Abel. 

10.  The  cubic  applied  to  angle  trisection.  —  By  higher  mathe- 
matics it  has  been  demonstrated  that   geometrical   problems 
which  can  be  solved  by  ruler  and  compass  correspond  alge- 
braically to  problems  whose  solution  can  be  effected  by  linear 
and  quadratic  equations  and  equations  reducible  to  quadratics, 
i.e.  by  equations  of  which  the  roots  will  involve  only  quad- 
ratic irrationalities  (square  roots,  and  square  roots  of  expres- 
sions involving   only  rational   quantities   and   square   roots). 
The  trisection  of  an  angle  is  a  type  of  geometrical  problem 
whose  solution  cannot  be  effected  with  ruler  and  compass ;  it 
is  possible  to  reduce  the  trisection  of  an  angle  to  an  algebraical 
problem,  the  solution  of  the  cubic  equation. 

Let  the  given  angle  which  is  to  be  trisected  be  denoted,  for 
convenience,  by  3  a.  Since  this  angle  is  given,  the  value  of  its 
sine  is  known.  If  the  angle  is  given  by  a  geometrical  drawing, 
the  ratio  of  the  perpendicular  h  dropped  from  a  point  at  a 
distance  r  from  the  vertex  on  one  side  to  the  second  side  to  r, 


400  UNIFIED  MATHEMATICS 

i.e.  -   gives  the  sine  of  the  angle.     Let  the  value  of  the  sine  of 
r 

the  given  angle  be  k. 

Given  sin  3  a  =  A;,  find  sin  «. 
sin  3  a  =  sin  (2  a  -f  a)  =  sin  2  a  cos  a  +  cos  2  a  sin  a 

=  2  sin  a  cos2  a  +  (cos2  a  —  sin2  a)  sin  a 
sin  3  a  =  3  sin  a  —  4  sin3  a. 
ft  =  3  sin  a  —  4  sins  a. 

This  equation  is  a  cubic  in  the  unknown  sin  a ;  for  convenience 
it  may  be  written  k  =  3  x  —  4  x3,  substituting  x  for  sin  a. 

There  are,  in  fact,  three  solutions  of  the  cubic  and  these 
three  solutions  correspond  to  the  fact  that  k  is  the  sine  not 
only  of  3  a,  but  also  of  180°  -  3  a,  and  n  •  360°  +  3  a,  and 
(2w  +  l)180°-3«. 

Thus  the  cubic  which  would  give  the  sine  of  10°,  trisecting 
the  angle  of  30°,  is  .5  =  3x  -  4a*»,  or  Ix3  -  3  x  +  .5  =  0.  The 
same  cubic  would  be  obtained  if  it  were  desired  to  trisect  the 
angle  of  150°,  or  of  390°,  or  750°,  •••.  There  are  an  infinite 
number  of  angles  which  have  this  same  sine,  .5,  but  there  will 
be  only  three  different  values  involved  when  the  sine  of  the 
third  part  of  each  of  these  angles  is  found.  In  the  equation 
4  x3  —  3  x  +  .5  =  0,  the  roots  represent  sin  10°,  sin  50°,  and 
sin  250°.  (See  problem  6,  page  396.) 

11.   Closer  approximation  to  located  roots.  —  The  method  will 
be  shown  by  a  numerical  illustration. 
The  equation 

(1)  a?-2z2-18a;  +  24  =  0, 

of  which  the  graph  is  given  on  page  71,  evidently  has  a  root 
between  4  and  5.  To  form  the  new  equation  whose  roots  are 
4  less  than  the  given  equation,  substitute  x'  +  4  for  x,  giving 

(2)  (x'  +  4)3  -  2(x'  +  4)2  -  18(«/+  4)  +  24  =  0. 
Assume  that  this  gives 

(3)  x's  +  Bxf*  +  Cxf  +  D=  0,  in  which  B,  C,  and  D  can  be 
obtained  by  expanding  and  combining  terms  in  (2).     The  left- 


hand  members  of  equation  (3)  and  equation  (2)  are  then  identi- 
cal. Evidently,  if  x  —  4'  is  substituted  for  x'  in  (2),  it  will  give 
the  original  equation,  and  consequently,  if  x  —  4  is  substituted 
in  (3),  it  will  give  the  original  equation.  Substituting,  we  have 

(4)  (x  -  4)3  +  B(x  —  4)2  +  C(x  —  4)  +  D  =  0,  which  is  identi- 
cal with  the  original  equation. 

If  the  left-hand  member  of  this  equation,  i.e.  the  original, 
is  divided  by  x  —  4,  the  remainder  is  D  and  the  quotient  is 
(x  —  4)2  +  B(x  —  4)  +  C ;  if  this  quotient  is  divided  by  (x  —  4), 
the  remainder  is  C ;  if  the  quotient  of  the  preceding  division 
is  divided  by  x  —  4,  the  remainder  is  B. 
x  -  4)  x3  -  2  x2  -  18  x  +  24  • 

+    * — +    **     ~^°  The   continued   division  by 

D    a? -4    is    effected    by    the 
+   4+24  \  . 

synthetic  process,  explained 


1  +   6   (+14  C 

in  section  4,  above. 

+    4 


B 

(5)  x'3  +  10  x'2  +  14  xr  —  16  =  0  is  then  the  equation  whose 
roots  are  4  less  than  the  roots  of  the  original  equation.  This 
should  be  verified  by  substitution  and  expansion.  The 
original  equation  has  a  root  between  4  and  5.  Hence  this 
equation  has  a  root  between  0  and  1.  By  trial  of  tenths,  .1, 
.2,  —  .9,  this  equation  is  found  to  have  a  root  between  .7  and 
.8.  Hence  the  original  equation  has  a  root  between  4.7  and  4.8. 

Form  the  new  equation  whose  roots  are  .7  less  than  the 
roots  of  (5). 

aj'  _  .7)1  +  10     +14        -  16       (.7 

+  .7)    +     .7  +   7.49  +  15.043 
1  +  10.7  +  21.49(-      .957 

+     .7  +    7.98 
1  +11.4(+  29.47 
.7 


(6)  z3  +  12.1  z2  +  29.47  z  -  .957  =  0. 


402 


UNIFIED  MATHEMATICS 


Equation  (5)  has  a  root  between  .7  and  .8 ;  hence  equation 
(6)  has  a  root  between  0  and  .1. 

By  trial  of  hundredths,  trying  .02,  .03,  .04  —  it  is  found  that 
this  equation  has  a  root  .03+,  between  .03  and  .04  and 
evidently  nearer  .03. 

Hence  our  original  equation  has  a  root  4.73+. 

In  this  way  we  can  compute  any  real  numerical  root  of  a 
rational  integral  algebraical  equation  to  any  desired  number 
of  significant  figures. 


ILLUSTRATIVE     PROBLEMS.  —  1.    Find     the    cube    root    of 
1,624,276  to  four  significant  figures. 


x3  —  1,624,276  =  0.     By  trial,  substituting   100,  200, 
found  to  have  a  root  between  100  and  200. 


x-100) 
100) 


for  x,  this  is 


x'  -  10) 
10) 


x"-7) 
7) 


1  +   0  +    0  -  1,624,276 
4-100  +10000  -1,000,000 

1  +  100  +  10000  (— 
+  100  +  20000 

624,276 

1  +  200  (  +  30000 
+  100 

1  +  300  -+  30000  - 
+  10  +  3100  - 

624,276 
331,000 

1  +  310  +  33100  (- 
+  10  +  3200 

293,276 

1  +  320  (+36300 
+  10 

1  +  330  +  36300  - 
+   7  +  2359  + 

293,276 
270,613 

1  +  337  +  38659  (- 
+   7  +  2408 

22,663 

1  +  344  (+  41067 

+  7 

1  +  351  +  41067  - 

22,663 

By  derivation,  the  roots 
are  100  less  than  the  roots  of 
(1 )  ;  hence  a  root  between  0 
and  100.  By  trial,  substitut- 
ing, root  between  10  and  20. 

By  derivation,  has  a  root 
between  0  and  10.  By  trial, 
a  root  between  7  and  8. 


By  derivation,  root  between 
0  and  1.  By  trial,  between 
.5  and  .6,  and  nearer  to  .5. 

Hence  the  root  of  the  original  is  117.5,  which  may  be  partially  checked 
by  four-place  logarithms. 


NUMERICAL  ALGEBRAIC  EQUATIONS          403 

2.    Compute  one  negative  root  of 

2 a;4  +  lOo3  -  8  a2  -  11  x  +  19  =  0. 
Negative  root  between  —  1  and  —  2. 


x  +  2)   2  +  10 
-2)     -  4 

-  8 
-12 

-  11     +19 
+  40     -  58 

2  + 

6 
4 

-20 
-  4 

+  29    (-  39 
+  48 

2  + 

2 
4 

-24 
+  4 

(+77 

+  77     -  39 
-  13.728  +  37.9632 

2- 

2 
4 

(-20 

-20 
J  -  2.88 

X  -  .6)  2  - 
+  .6)   + 

6 

l.i 

Boot  between  0  and 
1.       By    trial,     be- 
2  +    4.8    -  22.88    +  63.272    -  (1.0368    tween  .6  and  .7. 

+    1.2    -    2.16    -15.024 
2-    3.6    -25. 04  (+48. 248 

+    1.2    -    1.44 
2_    2.4  (-26. 48 

+    1.2 

2  -    1.2    -  26.48    +  48.248    —    1.0368    By  derivation  has  a 

root  betwen  0  and 
.1.  By  trial,  be- 
tween .02  and  .03. 

The  original  equation  has  a  root  —  2  +  .62+,  or  —  1.38~,  i.e.  between 
-  1.38  and  -  1.37. 

PROBLEMS 

See  the  two  preceding  sets  of  problems  and  use  the  results 
obtained. 

1.  Compute  to  three  significant  figures  the  largest  positive 
root  of  the  following  equations, 

d.  x*  -  2  x3  +  3  x2  -  18  x  +  21  =  0. 

2.  Compute  by  the  process  indicated  the  positive  root  of 
#2  _  3  x  —  21  =  0  to  three  decimal  places  ;  compute  the  same 
by  solving  as  a  quadratic,  and  compare  as  to  efficiency  the  two 
methods.  - 


404  UNIFIED  MATHEMATICS 

3.  Solve    the    equation    4^  —  3x-f-  .5=  0,   computing   the 
smallest  positive  root  to  four   decimal   places.     This   is   the 
value  of  sin  10° ;  check  by  your  table  of  sines. 

4.  In  problem  5,  page  245,  you  have  computed  the  sine  of 
3°  to  four  decimal  places.    Write  the  cubic  which  will  give  the 
sine  of  1° ;  compute  the  smallest  positive  root,  and  discuss  to 
what  decimal  place  it  could  be  carried  with  propriety  when 
the  sine  of  three  degrees  is  given  to  four  decimal  places. 

5.  Plot  the  graphs   of   the  two  equations  {        ~    '    2 

Note  that  the  points  of  intersection  give  the  solutions  of  the 
two  equations  regarded  as  simultaneous  ;  but  solving  the  two 
equations  as  simultaneous  equations,  we  are  led  by  substitution 
to  the  cubic  x  •  $  (z2  -  16)  =  1,  or  x3  - 16  x  -  4  =  0.  Solve  the 
cubic  and  compare  with  the  solutions  obtained  graphically. 

6.  Historical  problem.    The  great  Archimedes  proposed  the 
problem  to  cut  a  sphere  by  a  plane  in  such  a  way  that  the 
two  segments  of  the  sphere  should  have  to  each  other  a  given 
ratio.    Archimedes  showed  that  the  solution  could  be  obtained 
as  the  intersection  of  a  hyperbola  and  a  parabola.     If  the 
diameter  of  the  sphere  is  taken  as  10  and  k  as  the  ratio  of  the 
larger  to  the  smaller  segment,  this  problem  leads  to  the  cubic 

(k  —  11 

x3  —  300  x  +  2000  •> «•  =  0, 

k+I 

in  which  x  represents  the  distance  of  the  plane  from  the  center 
of  the  sphere.  Solve  to  two  decimal  places  when  k  =  2.  The 
plane  at  a  distance  x  from  the  center  then  trisects  the  sphere. 

7.  In  the  preceding  problem  show  that  the  solution  may  be 
obtained  as  the  intersection  of  a  hyperbola  and  a  parabola. 

8.  A  famous  problem  of  antiquity  is  the  problem  to  dupli- 
cate a  given  cube,  i.e.  to  solve  geometrically  x3  =  2  a3,  a  being 
the  side  of  the  given  cube.     Long  before  analytical  geometry 
was  invented  it  was  known  that  the  solution  could  be  given  as 
the  intersection  of  the  parabola  x2  =  ay  with  the  parabola 


NUMERICAL  ALGEBRAIC  EQUATIONS          405 

y*  =  2  ax.  Construct  the  graphical  solution  when  a  is  taken 
as  10. 

The  problem  may  also  be  solved  by  the  intersection  of  either 
of  the  two  given  parabolas  with  the  hyperbola  xy=2a2.  Verify. 

If  two  means,  x  and  y,  are  inserted  between  a,  and  2  a,  i.e. 

-  =  -  =  —,  then  x  is  the  solution  of  the  equation  x3  =2  a3. 
x  y  2  a 

This  method  reduced  the  problem  of  the  duplication  of  the 
cube  to  the  problem  of  inserting  between  two  given  numbers, 
or  lines,  two  geometric  means. 

9.    The  volume  of  a  spherical  segment,  greater  than  a  hemi- 
sphere, of  height  x  +  r,  is  given  by  the  expression 


o 

the  volume  of  a  sphere  is  f  Trr3.  Find  the  segment  of  a  sphere 
of  water  of  radius  10  which  will  be  equal  in  weight  to  a 
sphere  of  wood,  radius  10,  which  wood  is  only  .6  as  heavy  as 
water.  This  leads  to  the  cubic  equation 


or  x3  —  3  r>x  +  .4  r5  =  0, 

and  r  +  x  is  the  depth  to  which  the  sphere  of  wood  will  sink 

when  it  is  placed  in  water.      Compute  this  depth  when  r  =  10. 

10.  Ice  is  only  .92  as  heavy  as  water.  Use  the  equations  of 
the  preceding  problem,  substituting  .92  for  .6,  to  find  the  depth 
to  which  a  spherical  iceberg  of  radius  100  feet,  if  one  were 
possible,  would  sink  in  water. 


406 


UNIFIED  MATHEMATICS 


;Mi 


*\J  CS 

-III 

3 


z3  f  v  i^s 

• 


o    S 


o    o 


:    - 


CHAPTER   XXVI 

WAVE   MOTION 

1.  General.  —  In  nature  there  are  two   types   of   recurrent 
motion,  somewhat  closely  connected  mathematically,  in  which 
repetition  of  motion  occurs  at  regular  intervals. 

One  type  of  this  motion,  in  cycles  as  we  may  say,  repeats 
the  motion  in  one  place,  and  is  in  a  sense  stationary.  The 
tuning  fork  in  motion  moves  through  the  same  space  again 
and  again ;  a  similar  movement  is  the  motion  of  a  vibrating 
string.  Of  this  stationary  type  may  be  mentioned  the  heart- 
beats, the  pulse,  the  respiration,  the  tides,  and  the  rotation 
of  a  wheel  about  its  axis. 

The  second  type  of  recurrent  motion  transmits  or  carries 
the  vibratory  impulse  over  an  extent  of  space  as  well  as  time. 
The  waves  of  the  sea  are  of  this  character.  Sound  waves, 
electrical  vibrations  or  waves,  and  radiant  energy  vibrations 
are  transmitted  by  a  process  similar  to  that  by  which  the 
waves  of  the  sea  are  carried. 

Both  of  these  types  of  motion  are  representable  mathe- 
matically by  equations  involving  a  sequence  of  trigonometric 
functions.  To  the  fundamental  and  basic  function  involved, 
y  =  siii  x,  we  will  direct  our  attention  in  the  next  section  and 
to  simple  applications  in  other  sections  of  this  chapter. 

2.  The  sine  curve.  —  As  a  radius  vector  of  unit  length  ro- 
tates in  a  plane  with  uniform  velocity  about  a  center,  the  sine 
of  the  angle  6  fluctuates  between  1  and  —  1.     The  variation  of 
sine  6  may  be  represented  by  the   movement  on   the  y-axis 
of   the  projection  of  the  vector,  and  this  movement  of  the 

407 


408 


UNIFIED  MATHEMATICS 


Graph  of  y  =  sin  9  ;  a  pure  sinusoid 

The  length  AA'  equals  the  circumference  of  the  circle  ;  the  amplitude, 
vertical  distance  between  highest  and  lowest  points,  equals  the  diameter 
of  the  circle. 


Graph  of  y  =  sin  2  6;  a  sinusoidal  curve 
The  frequency  is  double  that  represented  in  the  preceding  graph. 


y  =  sin  8  +  sin  2  6;    obtained  by  addition  of  corresponding  ordinates  in 

the  two  preceding  curves 

This  type  of  curve  is  obtained  from  a  tuning  fork  having  an  octave 

overtone. 


WAVE  MOTION  409 

projection  is  termed  simple  harmonic  motion,  frequently 
abbreviated  S.  H.  M.  Precisely  the  same  type  of  movement 
is  given  by  the  projection  of  the  moving  vector  on  the  ovaxis, 
x  =  cos  6,  or  on  any  line  in  the  plane  of  the  motion, 

z  =  cos  (0  —  t), 

wherein  e  is  the  slope  angle  of  the  line  and  z  is  the  projection 
of  the  radius. 

If  the  vector  completes  one  revolution,  2irr,  or  360°,  in  1 
second,  the  period  of  the  motion  is  called  1  second,  and  the 
frequency,  or  the  number  of  repetitions  of  the  complete  move- 
ment or  cycle  in  a  second,  is  1  per  second.  If  the  complete 
revolution  is  effected  in  1  second,  the  period  is  £  second  and 
the  frequency  2  per  second.  The  graphs  of  y  =  sin  0  and  of 
y  =  sin  2  0  represent  under  these  conditions  the  progress  of  the 
ordinate  for  uniform  changes  in  0,  i.e.  for  uniform  changes 
in  the  time,  since  the  rotation  is  with  constant  angular  ve- 
locity. For  convenience  the  angle  is  conceived  as  measured 
in  radians  and  the  radius  is  taken  on  the  cc-axis  as  the  unit  to 
represent  one  radian  ;  the  abscissa  then  corresponds  either  to 
the  angle  measured  in  radians  or  to  the  length  of  arc  traversed 
by  the  end  of  the  moving  vector.  In  plotting  y  =  sin  0,  values 
of  0  from  0  to  360°  or  from  0  to  2  irr  are  plotted  on  the  horizon- 
tal or  0-axis.  Note  particularly  the  points  for  which  6=0, 
30°,  45°,  60°,  90°,  -  180°,  .»  360°  ;  or 


6  3     2 

Note  that  AA'  on  our  diagram  represents  one  complete  cycle 
or  period.  For  many  purposes  it  is  desirable  to  take  t,  the 
time  (in  seconds,  usually),  as  the  variable.  The  same  graph 
then  represents  y  =  sin  2  irt,  wherein  AA'  is  taken  equal  to  1  and 
the  horizontal  axis  is  the  Z-axis.  The  same  curve  represents 
y  =  sin  20  irt,  if  AA'  is  taken  as  T^  of  1  unit  of  time.  The 
upper  curve  in  our  diagram  is  a  pure  sinusoid,  the  distance  AA' 
representing  the  circumference  of  the  circle  of  which  the 
maximum  ordinate  is  the  radius. 


410 


UNIFIED  MATHEMATICS 


The  two  curves  plotted  should  be  carefully  studied ;  the 
lower  "curve  has  double  the  frequency  of  the  upper  and  one 
half  the  period.  The  swing,  amplitude  as  it  is  termed,  is  the 
same;  the  amplitude  is  the  algebraic  difference  between  the 

maximum  and  minimum  values  of 
the  function. 

Any  curve  representing 

y  —  a  sin  8   or   a  sin  *8 
or  a  sin  (kQ  -f  e) 

is  called  a  sinusoid.  We  shall  find 
that  the  graphs  of  y  =  a  cos  6, 
y  =  a  cos  k$,  and  y  =  a  cos  (kd  +  e) 
differ  from  the  preceding  only  in 
position. 

For  most  purposes  it  is  conven- 
ient to  plot  time  in  complete  units 
on  the  ordinary  coordinate  paper, 
the  unit  depending  on  the  period 
of  time  in  question.  For  a  com- 
plete rotation  in  one  minute  ten 
seconds  might  be  taken  as  one 
unit  on  the  horizontal  axis  with 
the  radius  as  vertical  unit,  and  the 
curve  would  differ  very  slightly 
from  our  curve.  The  highest  and 
lowest  points  would  fall  then  at 
15  and  45  respectively ;  0,  5,  7.5, 
10,  15,  20,  and  30  seconds  corre- 
spond then  to  0,  30°,  45°,  60°,  90°, 
120°,  and  180°  respectively. 

Physicists  and  engineers  com- 
monly draw  the  sinusoidal  curves 

which  are   of   frequent    occurrence    entirely   from   graphical 
considerations.     The    circle  with  the    desired   amplitude    is 


WAVE  MOTION  411 

drawn ;  the  angle  between  the  axes  is  bisected  and  re- 
bisected  (as  often  as  desired) ;  an  appropriate  length  for  a 
complete  cycle  is  taken  on.  the  horizontal  axis,  and  this  is 
divided  into  just  as  many  parts,  usually  16,  as  the  circum- 
ference of  the  circle  is  divided  by  the  axes  and  the  bisecting 
linesv which  were  drawn.  At  each  point  on  the  horizontal  axis 
an  ordinate  is  drawn  and  from  each  corresponding  point  on 
the  circle  a  horizontal  line  is  drawn  to  intersect  the  correspond- 
ing ordinate.  Corresponding  points  have  the  same  numbers 
if  on  the  circle  intersection  points  are  numbered  from  the 
right-hand  intersection  with  the  horizontal  axis  counter-clock- 
wise and  numbered  on  the  line  from  the  left-hand  end  of  the 
horizontal  length  taken  to  represent  the  time  of  one  cycle,  as 
indicated  on  our  diagram.  The  two  upper  figures,  page  408, 
were  drawn  by  this  method.  The  student  is  urged  to  make  both 
the  graphical  construction  and  the  construction  by  using  the 
numerical  values  of  the  sines  from  the  tables.  Compare  also 
the  work  under  Section  11,  Chapter  VII. 

PROBLEMS 

1.  Plot  the  curves  y  =  sin&°  and  y  —  sin  30°  on  the  same 
sheet  of  coordinate  paper ;  take  1  inch  as  radius  and  on  the 
horizontal  axis  take   1  inch  to   represent   60°.     For   a   pure 
sinusoid,  y  =  sin  x,  one  unit  on  x  should  be  the  length  of  the 
radius ;  then  3.14+  radians  represents  180°,  the  second  point 
in  which  the  curve  y  =  sin  x  cuts  the  axis  of  abscissas. 

2.  Plot  y  =  sin 2  wt ;  note  that  t  =  y1^,  y2^,  •••  corresponds  to 
36°  and  multiples ;  take  one  unit  for  t  as  6  times  the  radius 
chosen. 

3.  How  could  you  interpret  the  curve  of  the  preceding  exer- 
cise as  y  =  sin  4  wt  ? 

4.  Plot  10  points  of  y  =  sin  (&  —  30°).     This  curve  is  similar 
to  the  preceding;  it  is  30°  behind,  we  may  say,  the  regular 
sine  curve ;  the  "  lag "  is  30° ;  the  two  curves  y  =  sin  6  and 


412  UNIFIED  MATHEMATICS 

y  =  sin  (0  —  30°)  are  said  to  be  out  of  phase,  the  phase  angle 
of  the  second  being  —  30°.  The  "  phase  angle  "  is  of  particu- 
lar importance  in  the  theory  and  practice  of  alternating  cur- 
rents. 

5.  Plot  the  curve  E  =  110  sin  6.  Note  that  if  the  horizon- 
tal scale  be  taken  so  that  1  inch  represents  60°  and  the  vertical 
scale  such  that  1  inch  represents  110,  the  curve  is  precisely 
the  first  curve  of  problem  1.  This  curve  represents  the  vari- 
able electromotive  force  (e.  m.  f .)  developed  by  a  generator 
which  generates  a  maximum  e.  m.  f.  of  110  volts.  To  plot 
the  curve  no  knowledge  of  electricity  is  necessary,  but  com- 
plete interpretation  requires  technical  knowledge. 

3.  Sound  waves.  —  If  a  tuning  fork  for  note  lower  C  is  set 
to  vibrating,  the  free  bar  makes  129  complete,  back-and-forth, 
vibrations  in  one  second.  By  attaching  a  fine  point  to  the  end 
of  the  bar  and  moving  under  this  bar  at  a  uniform  rate,  as  it 
vibrates,  a  smoke-blackened  paper,  a  sinusoidal  curve  is  traced 
on  the  paper.  Our  curve  is  traced  by  a  bar  vibrating  50  times 
in  1  second. 


The  curve  y  =  sin  (50  X  2  irt) 
Tuning  fork  vibrations  recorded  on  smoked  paper. 

In  1  second  50  complete  vibrations  are  made ;  the  vertical 
distance  between  the  top  and  the  bottom  of  the  arcs  repre- 
sents the  distance  moved  by  any  point  on  the  moving  bar ; 
the  motion  is  simple  harmonic  (S.  H.  M.).  The  period  is  Jg- 
second ;  the  frequency  is  50 ;  the  amplitude  is  about  -£%  inch. 
If  the  smoked  paper  were  moved  with  uniform  velocity  under 
the  vibrating  bar  in  such  a  way  as  to  cover  50  times  the  cir- 
cumference of  a  circle  with  radius  JT  of  an  inch,  or  50  x  2  ?r  X  -fa 
inch  per  second,  the  curve  traced  would  be  almost  a  perfect 
sinusoid  of  the  type  y  =  sin  0.  The  points  move  of  course  on 


WAVE  MOTION 


413 


arcs  of  curves,  but  the  variation  from  a  straight  line  is  ex- 
tremely slight. 

Corresponding  to  each  movement  of  the  vibrating  rod  there 
is  a  movement  of  the  air.  As  the  bar  moves  to  the  right  it 
compresses  the  layer  of  air  to  its  right  and  that  compression 
is  immediately  communicated  to  the  layer  of  air  to  the  right ; 
as  the  bar  moves  back  and  to  the  left,  the  pressure  on  the  ad- 


Vibration  records  produced  by  the  voice 

"  a"  as  in  "  ate  "  ;  "ou  "  as  in  "about  "  ;  "  r  "  in  "relay  ";  "6" 
in  "be";  and  "a"  in  "father."  The  tuning  fork  record,  frequency 
60  per  second,  gives  the  vibration  frequencies. 

jacent  air  is  released  and  a  rarefaction  takes  place.  In  ^  of  1 
second  you  have  the  air  adjacent  to  the  rod  compressed,  back 
to  normal,  and  rarefied;  during  this  time  the  neighboring  air 
is  affected  and  the  compression  is  communicated  a  distance 
which  is  the  wave  length  of  this  given  sound  wave.  In  1  second 
this  disturbance  is  transmitted  1100  feet  at  44°  Fahrenheit. 
The  wave  length  for  this  sound  wave  then  is  it|p-  =  22  feet. 

The  wave  length  is  commonly  designated  by  X.  If  v  is  the 
velocity,  and  t  the  time  of  one  vibration,  X  =  vt. 


414  UNIFIED  MATHEMATICS 

The  notes  of  the  key  of  C  on  the  natural  scale  have  the  fol- 
lowing vibration  frequencies : 

cdefgabc' 
256  288  320  341.3  384  426.7  480  512 

The  intensity  or  loudness  corresponds  in  the  rod  to  the 
length  of  swing  of  the  vibrating  rod ;  as  this  amplitude  de- 
creases, the  intensity  of  the  sound  decreases.  For  small  am- 
plitudes the  vibratory  motion  gives  a  convenient  way  of 
measuring  small  intervals  of  time. 

Thus  on  the  above  diagram  if  the  tone  of  the  note  lower  C, 
y  =  sin  256  irt,  were  represented,  each  complete  wave  would 
represent  T^  of  1  second ;  each  half  or  each  arch  would  rep- 
resent YST  °f  1  second.  Tuning  bars,  with  periods  J^  and 
-j-^  of  1  second  are  run  electrically  for  timing  purposes. 

The  curve  y  =  sin  6  4-  sin  2  6  represents  the  combination  in 
sound  of  two  tones  which  differ  by  an  octave.  Precisely  the 
type  of  curve  which  is  represented  by  our  diagram  can  be 
produced  mechanically  by  the  record  of  a  vibrating  tuning 
fork1  which  sounds  not  only  the  principal  note  but  also  the 
octave  overtone,  due  to  the  fact  that  the  bar  vibrates  about  the 
middle  point  at  the  same  time  that  it  vibrates  about  the  end. 
Vibrating  strings  also  have  multiple  vibration,  overtones  and 
other  tones.  Harmony  is  the  result,  in  general,  when  the 
vibrating  instrument  gives  vibrations  which  are  connected 
with  the  fundamental  vibration  by  simple  numerical  relations, 
like  that  of  the  overtone. 

Thus  the  notes  of  the  major  chord,  key  of  C,  c,  e,  g,  c,  on 
the  piano,  have  the  vibration. frequencies  in  the  ratios  4  to  5 
to  6  to  8. 

4.  Helical  spring.  —  Similar  to  the  vibrations  of  the  air  are 
those  of  a  spiral  wire  spring  which  oscillates  back  and  forth 
when  a  weight  is  suspended  by  the  spring;  the  successive 
compressions  and  elongations  of  the  wire  correspond  quite 

1  See  Miller,  The  Science  of  Musical  Sounds,  p.  188,  for  photograph. 


WAVE  MOTION  415 

closely  to  the  condensations  and  rarefactions  of  the  air.  The 
position  of  the  weight  at  any  instant  can  be  given  by  an 
equation  entirely  similar  to  the  equation  above  of  note  C. 
Thus  if  the  time  of  one  complete  vibration  is  ^  second,  and 
the  maximum  displacement  is  4  inches,  the  equation  is 

y  =  4  sin  4  irtr, 

This  gives  the  elevation  above  and  below  the  point  at  which 
the  weight  comes  to  rest.  Perfect  elasticity  of  the  spring  is 
assumed. 

5.  Light  waves.  —  Light  waves  have  a  much  higher  velocity 
than  sound  waves,  300  x  lO6  meters  per  second.     The  different 
wave  lengths  correspond  to  different  colors,  just  as  different 
wave  lengths  in  sound  waves  correspond  to  different  tones. 
The  wave  length  of  the  light  from  burning  sodium  (Z),  of  the 
spectrum)    is  0.5890  x  10~6  meters  per  second,  and  for  other 
colors    varies    for    the    visible    spectrum    between   .39    and 
.75  x  10"6  meters.     The  vibration   frequency   of   the   sodium 
light  is  the  number  of  these  waves  which  occur  in  one  second 
of  time,  hence  since  these  waves  cover  300  X  106  meters  in  one 
second  the  frequency  n  is  such  that 

n  -  A.  =  v,  or  n  •  0.589  x  10~6  =  300  x  106, 

300  x  106 
whence    n  = — -  =  509  x  1012    vibrations    per    second. 

Radiant  energy  is  of  the  saine  general  nature  with  longer 
waves.  Light  waves  differ  from  the  sound  waves  in  having 
transverse  vibrations,  not  longitudinal. 

6.  Electricity.  —  In  electricity,  particularly  in  the  discussion 
of  alternating  currents,  the  sine  curve  plays  a  prominent  role. 

The  equations  e  =  156  sin  0, 

i  =  4  sin  0, 

and  p  =  ei  =  624  sin2  0  =  624(£  -  1  cos  2  6) 

=  312  -  312  cos  2  0, 


416 


UNIFIED  MATHEMATICS 


represent  respectively  the  electromotive  force,  e,  measured  in 
volts,  and  the  current,  i,  measured  in  amperes,  and  the  power, 

p,  measured  in  watts 
of  an  ordinary  electric 
current. 

In  general,  current 
and  electromotive 
force  are  "out  of 
phase " ;  the  equa- 
tions when  the  cur- 
rent lags  30°  behind 
the  electromotive 
force  are, 


e  =  156  sin  0, 

i  =  4  sin  (0-30°). 


Sinusoids  traced  by  electrical  means 

Oscillogram  of  an  alternating  current  in  which 

current  and  e.  m.f.  are  "  in  phase." 

On  the  two  diagrams 

we  have  represented  by  a  photographic  process  the  magnitude 
of  the  current  and  electromotive  force  of  an  alternating  cur- 
rent. The  current  is  represented  by  the  curve  with  the 
smaller  amplitude. 
In  the  first  illus- 
tration current  and 
e.  m.  f.  are  "  in 
phase,"  and  under 
these  conditions  a 
maximum  of  power 
is  developed ;  in  the 
second  illustration 
current  and  e.  m.  f. 
are  "  out  of  phase," 
the  current  lagging 
behind  the  e.  m.  f . 

The  power  at  any  instant  delivered  by  an  alternating  current 
is  given  by  the  product  of  the  current  and  the  e  .m.  f .  at  that 
instant.  Employing  the  formulas, 


Oscillogram  showing  current  curve  (lower)  lag- 
ging 90°  behind  e.  m.  f.  curve 


WAVE  MOTION  417 

cos  (a  —  ft)  —  cos  a  cos  ft  -f  sin  a  sin  /?, 
cos  (a  +  ft)  =  cos  «  cos  /3  —  sin  a  sin  ft,  whence 
cos  (a  —  ft)  —  cos  (a  -f  ft)  =  2  sin  a  sin  ft, 
show  that p=  ie  sin 0 sin (0  —  30°)  may  be  reduced  to 

—  [cos  30°  -  cos  2(6  -  15°)]. 
tt 

Plot  the  curve  showing  the  power  at  any  instant,  when 
e  =  156  cos  0  and    i  =  4  cos  (0  —  30°). 

Note  that  this  power  curve  is  also  a  sinusoidal  curve  but  placed 
with  reference  to  a  horizontal  line  which  runs  270  units  above 
the  x-axis. 

PROBLEMS 

1.  Plot    the    curves    y  =  sin  256  IT  f  and    y  =  sin  512  w  ft 
using  ^  inch  for  1  on  the  vertical  axis  and  6  half-inches  for  yi-g- 
of  1  second  on  the  t  or  horizontal  axis.     Treat  the  equations 
as  y  =  sin  2  irtr,  and  y  =  sin  4  Trf,  respectively,  substituting  for 
t,  0,  .1,  .2,  .3,  — ,  .9,  and  1  instead  of  -fa  of  T^-g-,  T2^  of  T^-g-,  •••. 

Note  that  the  unit  T|?  taken  as  1  on  the  horizontal  axis,  disposes  of  the 
difficulty  of  the  awkward  fractions. 

2.  What  is  the  frequency  of  the  vibrations  in  the  curves 
of  the  preceding  example  ?     What  are  the  corresponding  wave 
lengths  ? 

3.  How    would    y  =  cos  2  irtr   differ    from    the    curve    for 
y  =  sin  2  -rrV  ?  f  Write  10  values  of  y  =  cos  2  irf  for  t  =  0,  -fa, 

i>  i>  ¥'  •?'  TZ>  i'  '"  ** 

Note  that  these  angles  correspond  to  0°,  30°,  45°,  60°,  •••  respectively. 

4.  Use    the   equation    cos  9  =  sin  (90°  +  6)    to    show   that 
y  =  sin  0  lags  90°  behind  y  =  cos  6. 

5.  Draw  the  graphs  of  y  =  cos  0  and  y  =  cos  2  0;  divide  the 
arc  of  the  circle  into  24  equal  parts  and  take  the  distance  rep- 
resenting 2  TT"  as  divisible  by  24. 


418  UNIFIED   MATIIKMATICS 

6.  Draw  the  graph  of  y  =  sin  (  *  +  0  )    and  compare  with 

V4        / 
y  =  sin  6.     Discuss  the  corresponding  motion  of  the  moving 

point  on  the  vertical  axis. 

7.  The  limits  of  hearing  are  for  vibrations  of  16  per  second 
and  40,000  per  second.     What  are   the  corresponding  wave 
lengths  ? 

8.  Plot  on  the  same  diagram  the  two  curves, 

e  =  156  sin  0, 

i  =  4  sin  (0  -  30°). 

9.  In  problem  8  find  the  value  of  e  for  each  30°  to  360°. 
This  completes  a  "  cycle  "  of  values.     The  time  of  this  move- 
ment in  a  60-cycle  system  is  ^  of  1   second.     What  is  the 
value  of  t  for  the  angles  given,  and  also  for  6  =  45°,  135°,  225°, 
and  315°  ? 

10.  On   the   curve,   on  the   same   axes    as   the   preceding, 
i  =  4  sin  (6  —  30°),  read  the  values  of  Q  for  the  angles  30°,  45°, 
60°,  90°,  •••  to  360°.     These  may  represent  current  in  the  cir- 
cuit of  problems  5  and  6 ;  the  current  lags  30°  behind  the 
e.  m.  f.     What  interval  of  time  is  represented  by  the  30°  lag  ? 

11.  Plot  to  the  same  axes  the  curves,  i  =  4  sin  (6  +  40°), 

e  =  156  sin  0. 

The  curve  of  i  here  leads  the  curve  of  e  by  40°. 
In  the  case  of  i,  what  are  convenient  values  of  6  to  plot 
without  using  tables  ? 

12.  Assuming  that  it  takes  -^  of  1  second   for  one  com- 
plete cycle  of  i  or  e  in  problem  8,  find  the  time  difference 
represented  by  the  40°  angular  difference.     Find  angles  ap- 
proximately corresponding  to  ^  ^,  T^,  T^,  and  ^  of 
1  second. 

7.    Sine  curve;  circle;  ellipse;  cylinder.  —  If  a  circular  cyl- 
inder, such  as  the  one  in  our  diagram,  is  cut  by  any  plane,  the 


WAVE   MOTION 


419 


intersection  is  an  ellipse. 
Thus,  the  plane  through 
AOB  in  our  diagram,  in- 
clined at  an  angle  of  45°^ 
cuts  the  cylinder  in  an 
ellipse  whose  equation  is 

#     JP_t 

a?     2  a? 


1. 


The  circular  base  and  any 
parallel  section  has  the 
equation 

O2"f"o2  = 

Taking  the  portion  of  the 
cylindrical  surface  be- 
tween the  elliptical  curve 
and  the  circular  curve 
through  the  same  center 
and  unrolling  it  gives  a 
pure  sinusoid, 

y  =  a  sin  6. 

In  the  figure  PQ  =  QM, 
since  the  cylinder  is  cut 
at  an  angle  of  45°.  But 
QM,  the  ordinate  on  the 
circle,  equals  a  sin  9  and 

r 


Sinusoid  developed  by  means  of 
cylinder 


A  plane  intersecting  all  the  elements  of 

a  circular  cylinder  cuts  the   surface 

in  an  ellipse 

the  arc  AQ,  which  will  be  the  ab- 
scissa, is  a6. 

Hence  the  curve,  when 
rolled  out,  is 

y  =  a  sin  6 

and  will  give  a  complete 
arch  of  the   sinusoid  if 

a  circular       u?Per  and  lower  Portions 
of  the  surface  are  given. 


420 


UNIFIED  MATHEMATICS 


Experimentally  the  student  can  develop  this  curve  by  roll- 
ing a  sheet  of  paper  about  a  cylinder  and  cutting  out  with  a 
sharp  knife  the  required  portions. 


Piston-rod  diagram 
AB,  the  stroke  ;  HC,  connecting  rod  ;  0(7,  crank  arm. 

8.  Piston-rod  motion.  —  The  common  piston-rod  motion  of 
engines  furnishes  abundant  trigonometric  material,  much  of 
which  is  of  sufficiently  elementary  character  so  that  by  the 
application  of  simple  formulas  problems  of  interest  to  the 
engineer  can  be  solved. 

The  essential  features  for  our  purposes  are  the  piston  head 
H,  the  connecting  rod  HO  of  length  I,  the  crank  arm  OC  of 
length  r,  and  the  stroke  AB,  which  is  the  distance  through 
which  the  piston  head  H  moves.  Were  the  connecting  rod 
infinite  in  extent,  the  motion  of  H  would  be  simple  harmonic 
motion  when  C  is  rotating  with  uniform  velocity  about  0. 

In  modern  engines  the  ratio  of  I  to  r  varies  from  3  to  1,  low, 
to  4.8  to  1,  which  is  approximately  that  of  a  Ford  engine.  It 
is  desired  to  find  for  each  position  of  the  piston  head  the 
angle  a  of  the  connecting  rod,  the  angle  0  of  the  crank  shaft, 
and  also  the  effective  pressure,  called  the  tangential  compo- 
nent, of  the  connecting  rod  to  turn  the  crank  shaft. 

In   the   first  place,  when  I :  r  =  4.8 : 1,   the   angle  a  never 

exceeds  arc  sin  — .    Determine  this  angle  in  degrees.     As  the 
4.8 

pressure  P  at  H  is  horizontal,  only  a  portion,  Pcos  a,  of  this 


WAVE  MOTION 


421 


pressure  is  communicated  to  the  connecting  rod.  Discuss  the 
variation  in  pressure  due  to  the  inclination  of  the  connecting 
rod  and  note  that  it  is  relatively  small.  Of  course  the  pressure 
of  the  gas  in  the  cylinder  chamber  is  not  uniform  and  this 


Connecting  rods  and  cranft  arms  in  six-cylinder  automobile  engine : 
ratio  I :  r  =  3 : 1 

variation  is  much  more  serious  than  the  variation  due  to  the 
angle  of  a  connecting  rod. 

Find  also  the  maximum  vertical  pressure  P  sin  a  on  the 
cross-head  support. 

Show  that  the  length  of  the  stroke  AB  is  equal  to  the 
diameter  of  the  crank  circle. 

The  position  of  the  pfston  head  is  indicated  in  decimal  parts 
of  the  total  stroke  2  r,  as  measured  from  A.  In  our  diagram 
the  piston  head  is  at  .75  of  the  stroke.  To  determine  a  and  0 
when  the  position  of  the  piston  head  is  given  we  solve  a 
triangle  in  which  the  three  sides  are  known.  Thus  on  our 
diagram  O0'=  4.8 ;  HO  =  4.3 ;  HC=  4.8  ;  and  0(7=  1.  Solve 
this  triangle  and  determine  the  angles  a  and  6.  Commonly  a 
diagram  is  drawn  in  which  the  angles  6  of  the  crank  arm  are 
plotted  as  abscissas  and  the  piston  displacements  are  plotted 
as  ordinates. 


422 


UNIFIED  MATHEMATICS 


PROBLEMS 

1.  Given  0  =  0,  10°,  20°,  30°,  60°,  90°,  120°,  150°,  and  180°, 
find  the  corresponding  values  of  a  and  the  positions  of  H  as 
decimal  parts  of  the   stroke.     Plot   angles   as   abscissas   and 
piston  displacements  as  ordinates.     How  could  you  complete 
this  to  360°  ?     Could  you  go  on  beyond  360°  ? 

2.  Assuming  that  the  piston  rod.  transmits  a  uniform  pres- 
sure  of  200   lb.,   find   the    effective    turning   pressure   when 
0  =  30°,  60°,  and  90°.     Resolve  the  force  at  C  into  two  com- 
ponents, one  normal  to  the  crank  shaft  and  the  other  along 
the  crank  shaft.    The  component  normal  to  the  crank  shaft,  the 
tangential  component,  is  effective,  i.e.  does  the  work.     The 
radial   pressure  is  also  computed   and   is  used   to   determine 
friction  loss.     Find    the  values  of   the  radial   pressure  corre- 
sponding to  above  tangential  pressure. 

Compute  in  this  case  for.  the  given  angles  the  pressure 'on 
the  cross-head  support  by  the  connecting  rod  due  to  the  com- 
ponent, P  sin  a. 


Connecting  rod,  crank  arm,  and  cylinder  on  A.  T.  &  S.  F.  locomotive  1444 
The  stroke  is  30  inches  and  connecting  rod  60  inches. 

3.  Draw  the   graph   illustrating   relative   positions  of   the 
piston  head  and  the  connecting  rod  when  the  crank-pin  is  at 
the  lowest  point,  in  the  locomotive  illustrated  above. 

4.  Find  the  number  of   strokes  of   the  piston  per  minute 
when  the  train  moves  60  miles  per  hour,  given  that  the  driving 
wheels  are  57  inches  in  diameter. 


CHAPTER   XXVII 

LAWS    OF    GROWTH 

1.  Compound  interest  function.  —  The  function  S  =  P(l+  i)n 
is  of  fundamental  importance  in  other  fields  than  in  finance. 
Thus  the  growth  of  timber  of  a  large  forest  tract  may  be  ex- 
pressed as  a  function  of  this  kind,  the  assumption  being  that 
in  a  large  tract  the  rate  of  growth  may  be  taken  as  uniform 
from  year  to  year.     In  the  case  of  bacteria  growing  under 
ideal  conditions  in  a  culture,  i.e.  with  unlimited  food  supplied, 
the  increase  in  the  number  of  bacteria  per  second  is  propor- 
tional to  the  number  of  bacteria  present  at  the  beginning  of 
that  second.      Any  function  in  which  the  rate  of  change  or 
rate  of  growth  at  any  instant  t  is  directly  proportional  to  the 
value  of  the  function  at  the  instant  t  obeys  wha^t  has  been 
termed  the  "  law  of  organic  growth,"  and  may  be  expressed  by 
the  equation, 

y  =  cekt, 

wherein  c  and  k  are  constants  determined  by  the  physical  facts 
involved,  and  e  is  a  constant  of  nature  analogous  to  TT.  The 
constant  k  is  the  proportionality  constant  and  is  negative  when 
the  quantity  in  question  decreases ;  c  is  commonly  positive ; 

e  =  2.178  .». 

The  values  of  the  function  of  x,  cekx,  increase  according  to 
the  terms  of  a  geometrical  progression  as  the  variable  x  increases 
in  arithmetical  progression. 

2.  IT  and  e.  —  A  function  can  be  found  by  methods  of  the 
calculus  which  is  such  that  the  rate  of  growth  of  the  function 

423 


424 


UNIFIED  MATHEMATICS 


at  any  instant  t,  or  x,  exactly  equals  the  value  of  the  function  at 
that  instant.     This  function  is  given  by  the  equation, 

y  =  ef  or  y  =  e?. 
The  constant  e  is  represented  by  the  series 


wherein  [3,    ( 
[4  =  4x3x2 
teger,  represei 
down  to  1. 
2.71828;  to  1( 
The  curve 
equation 

is  such  that  a 
curve   the    sic 
merical    value 
that  point. 
The  graph 
that  the  slop< 
the  negative 
that  point. 
The  graph 
that  the  slope 

ailed    factorial    3,   represents    3x2x1,    and 
X  1,  and,  in  general,  \n,  n  being  a  positive  in- 
its  the  total  product  of  all  the  integers  from  n 
The  sum  of  this  series  to  5  decimal  places  is 
)  places,  =  2.7182818285. 
represented  by  the    .  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  ,  

±t---- 

y  =  ex 
b  any  point  on  this 

>pe   equals    in  nu- 
s    the   ordinate    at 

^  "     ;:  ~JtlT~ 

•P                              I      '  c-      c-ttn'h           •- 

1            i 

4-  -  f-  -  -  - 

•"  at   anv   point  is                  -  -  4* 

of  the  ordinate  at 

/           f 

1        ,1  ^_ 

n                                                             1                 '  \-~ 

I    1   / 

oi  y  —  e**  is   such 

/ 

;•          L        •              1                                                                                     i 

' 

at  any  point  ISA;     ±  :  :  : 

j   f                                        rr 

:::::::::::!? 

1  l!|!l       IIJJU  II  |||lll4^M 

1                "^ 

^••M  •>•••  U-I-I-L 

::|::::::::|!!     rrrt         :  0: 

_£  j: 

^  3:±--  

r=!=;4:::^^S5 

Graphs  of  y  =  6?  and  y  =  1(H 

times  the  corresponding  value  of  the  function  at  that  point. 
Values  of  the  .function  y  =  e*  may  be  determined  by  loga- 
rithms. 


LAWS  OF  GROWTH  425 

Thus  to  find  the  points  on  y  =  e*,  for  which 

x  =  —  2,  —  1,  0,  .1,  .2,  .5,  .8,  1,  2,  and  3  respectively, 
we  take  first  log  y  —  x  log  e ;  since  log  e  =  .4343 

logy  =  -.8686,  y=  .135 

=    9.1314-10 

log  y  =  -  .4343,  y  =      .368 

=    9.5657-10 

log  ?/  =    0,  y  =  1 

logy=      .0424,  y=  1.103 

log#=      .0869,  y  =  1.222 

logy=      .2171,  y=  1.649 

logy=      .3474,  y=  2.225 

logy  =      .4343,  2/=  2.718 

logy=      .8686,  y=  7.390 

a?=      3,  logy=    1.3029,  y  =  20.09 

Similarly,  if  y  —  cekx,  log  y  =  log  c  +  kx  log  e,  and  these  values 
are  obtained  by  logarithms. 
The  limit  of  the  expression 


as  n  approaches  infinity  gives  the  value,  e.  When  n  is  taken 
as  a  large  positive  integer,  it  can  readily  be  shown  that  this 
expression 


has  a  value  differing  but  slightly  from  e. 

e  and  IT  may  be  called  fundamental  constants  of  nature ;  in 
mathematical  work  as  applied  to  statistics  and  to  physical 
problems  of  varied  kinds  these  constants  often  appear. 

3.  Natural  logarithms.  —  The  first  logarithms  as  computed 
by  Napier  were  not  calculated  to  any  base,  but  were  founded 


426 


UNIFIED  MATHEMATICS 


upon  the  comparison  between  an  arithmetical  and  a  correspond- 
ing geometrical  progression.     However,  a  base  of  the  Napierian 

logarithms   can  be   established,  and   it   is   approximately  -. 

e 


Graph  of  y  =  log«x 

The  ordinates  represent  the  "natural  logarithms" 
of  the  numbers  represented  by  the  abscissas. 

From  the  mathematical  point  of  view,  and 
indeed  for  many  applications  of  mathe- 
matics to  physical  problems,  the  base  e  is 
preferable  to  10  as  the  base  of  a  system  of 
logarithms.  Logarithms  to  the  base  e  are 
called  natural  logarithms. 


4.  Application.  —  The  most  immediate  application  of  a 
function  in  which  the  growth  is  proportional  to  the  function 
itself  is  to  the  air.  The  decrease  in  the  pressure  of  the  air 
at  the  distance  h  above  the  earth's  surface  is  proportional 

to  h. 

h_ 

The  expression  P  =  760  e  7990  gives   the  numerical  value  of 

the   pressure  in   millimeters  of  mercury  for  7t  measured  in 
meters.     The  negative  exponent   indicates  that   the  pressure 


LAWS  OF  GROWTH  427 

decreases  as  h  increases.  In  inches  as  units  of  length  of  the 
mercury  column,  h  in  feet, 

P  =  29.92  e~^. 
This  is  known  as  Halley's  Law. 

The  growth  of  bean  plants  within  limited  intervals  and  the 
growth  of  children,  again  between  quite  restricted  limits, 
follow  approximately  the  law  of  organic  growth.  Radium  in 
decomposing  follows  the  same  law,  the  rate  of  decrease  at  any 
instant  being  proportional  to  the  quantity.  In  the  case  of 
vibrating  bodies,  like  a  pendulum,  the  rate  of  decrease  of  the 
amplitude  follows  this  law ;  similarly  in  the  case  of  a  noise 
dying  down  and  in  certain  electrical  phenomena,  the  rate  of 
decrease  is  proportional  at  any  instant  to  the  value  of  the 
function  at  the  instant. 

PROBLEMS 

1.  By  experiment  it  has  been  found  that  1000  of  the  so- 
called  "hay  bacteria"  double  their  number,  under  favorable 
conditions,   in   20   minutes.     Find    the    rate    of   growth    per 
minute.     Take  n  =  1000  eu, 

and  determine  Jc  by  substituting  n  =  2000,  t  =  20,  and  solving 
for  Tc.  Determine  the  number  that  would  grow  from  1000 
bacteria  in  1  hour  ;  in  1  day. 

2.  The  cholera  bacteria  have  been  found,  under  favorable 
conditions,  to  double  their  number  in  30  minutes.     Determine 
the  rate  of  growth  per  minute,  and  the  number  that  would 
grow  in  one  day  from  1000. 

NOTE.    The  favorable  conditions  cannot  be  continued  for  such  a  period. 

3.  Assuming  that 

P=760e~i^ 

find  the  value  of  h  which  will  reduce  the  pressure  of  the  air  by 
1  mm.  Take  the  logarithm  of  both  sides  and  note  that 


428  UNIFIED  MATHEMATICS 

-  Iog10  e  must  reduce  log  760  to  log  759.     Find  at  what 
8000 

height  the  mercury  column  is  reduced  1  cm.  At  what  height 
would  the  pressure  be  reduced  to  660?  Are  these  heights 
ever  attained  ? 

4.  Find  the  barometric  pressure  when 

h  =  1000,  5000,  10,000,  and  15,000  feet,  assuming 

P=  29.92,  when  h  =  0,  in  which  P  is  the  numerical  value 

of  the  pressure  in  inches  of  mercury  and  the  height  is  h  feet. 

Find  the  height  for  which  the  pressure  decreases  ^  of  1  inch. 

5.  Show  that  if  the  height  of  the  elevation  is  measured  in 
miles,  the  pressure  in  inches  is  given  approximately  by  the 

formula  A 

P  =  29.92  e  i. 

Note  that  26,000  is  nearly  5  x  5280.     The  constant  26,200,  above,  is 
taken  for  simplicity  instead  of  26,240. 

6.  To  what  change  in  height  does  the  maximum  variation 
of    the    barometer    recorded    on    the    photograph,,    page    60, 
correspond  ? 

7.  Compute    by    the    progressive    method    of    Section    4, 
Chapter  XII,  the  value  of  e,  from  the  series, 


by  summing  8  terms  to  8  decimal  places. 

8.  Plot  on  the  same  diagram  and  compare  the  two  graphs,  of 

y  =  e1  and  y  =  10*. 

9.  Plot  the  curve  y  =  e~z*,  taking  ^  inch  as  .1  on  the  hori- 
zontal and  on  the  vertical  axes.     This  curve  represents  what 
is  termed  the  normal  distribution  curve,  which  is  of  funda- 
mental importance  in  all  statistical  work.     In  general,  large 
groups  of  individuals  may  be  distributed  as  to  ability  in  any 
given  quality  over  the  area  under  such  a  curve  ;  the  middle  ab- 
scissa at  x  =  0  represents  average  ability,  and  deviation  to  one 
side  or  the  other  represents,  on  one  side,  ability  above  the  aver- 


LAWS  OF  GROWTH 


429 


age  and,  on  the  other,  ability  below  the  average.  The  total 
number  of  all  individuals  considered  is  represented  by  the 
area  between  the  curve  and  the  avaxis. 

An  interesting  graphical  illustration  of  a  normal  distribu- 
tion curve  is  the  crowd  at  a  football  game  when  a  great 
bleacher  is  not  filled.  The  central  aisles  are  all  tilled  to  a 
height  representing  the  middle  ordinate  and  from  this  out  in 
either  direction,  the  ordinates  drop  off,  frequently  in  strikingly 
symmetrical  manner,  and  corresponding  quite  closely  to  the 
normal  distribution  curve. 

5.  The  curve  of  healing  of  a  wound.  —  Closely  allied  to  the 
formulas  expressing  the  law  of  organic  growth,  y  =  ekt,  and  the 
law  of  "  organic  decay,"  y  =  e~kt,  is  a  recently  discovered 
law  which  connects  algebraically  by  an  equation  and  graph- 
ically by  a  curve,  the  surface-area  of  a  wound,  with  time 

Sg.rm.         Scofe-- „          Wound  •      '      • 


30  15^--- 

_ 

:H:::::::: 

:::::::::::::::::::5 

i£t::^±t£  +  -Jj)::: 

20  10  :: 

^V 

;;:; 

1        c^ 

^v^^ 

10    5  3f  - 

>           ; 

"!•!>•! 

|:j 

._.:*• 

=  |!Si!;|!::i:E::|:::: 

DayO       4        8       12      16     20     24     28     32     36     40     44      48     52     56      GO     6466 
MaylS    J9     23      27     31Je.4       8      12     16     20     24     28./Z.2      6      .10     14     1820 

Progress  of  healing  of  a  surface  wound  of  the  right  leg,  patient's  age  31  years 
The  observed  curve  oscillates  about  the  smoother,  calculated  curve. 

expressed  in  days,  measured  from  the  time  when  the  wound  is 
aseptic  or  sterile.  When  this  aseptic  condition  is  reached, 
by  washing  and  flushing  continually  with  antiseptic  solutions, 
two  observations  at  an  interval  commonly  of  four  days  give 
the  "index  of  the  individual,"  and  this  index,  and  the  two 
measurements  of  area  of  the  wound-surface,  enable  the  physi- 
cian-scientist to  determine  the  normal  progress  of  the  wound- 
surface,  the  expected  decrease  in  area,  for  this  wound-surface 
of  this  individual.  The  area  of  the  wound  is  traced  carefully 


430 


UNIFIED   MATHEMATICS 


on  transparent  paper,  and  then  computed  by  using  a  mathe- 
matical machine,  called  a  planimeter,  which  measures  areas. 

The  areas  of  the  wound  are  plotted  as  ordinates  with.the 
respective  times  of  observation  measured  in  days  as  abscissas. 
After  each  observation  and  computation  of  area  the  point  so 


Sg.rm. 

10 

V 

- 

\ 

^ 

^. 

^, 

' 

• 

I 

„/ 

,•/, 

Ht 

II 

„,. 

X' 

•/.  ll 

\ 

^ 

** 

" 

r 

-: 

/ 

•—  — 

-•^ 

s 

\ 

.' 

-N 

i 

- 

/ 

•"- 

- 

„  ~ 

\ 

\ 

j*-*' 

•£ 

-• 

s 

s 

- 

•>, 

* 

- 

\> 

/ 

S 

S 

s~ 

^ 

s 

? 

V 

•^, 

•f 

-" 

— 

-—  —  ^ 

o 

•j 

* 

— 

'  —  , 

-^ 

2te 

^M 

^-~ 

I'll, 

-<  / 

/•'•< 

1 

\ 

\ 

•2 

i 

a 

2 

t 

) 

16 


•24 


32 


Sept.  23         Oct.  1  9  17  25        Nov.  2  8       12 

Progress  of  a  surface  wound  of  the  right  knee 
Two  infections  in  the  course  of  healing  are  indicated. 

obtained  is  plotted  to  the  same  axes  as  the  graph  which  gives 
the  ideal  or  prophetic  curve  of  healing.  Two  such  ideal 
curves  and  also  the  actual  observed  curves  are  represented  in 
our  diagrams. 

When  the  observed  area  is  found  markedly  greater  than 
that  determined  by  the  ideal  curve,  the  indication  is  that 
there  is  still  infection  in  the  wound.  This  is  the  case  de- 
picted, as  will  be  noted,  in  the  smaller  diagram.  A  rather 
surprising  and  unexplained  situation  occurs  frequently  when 
the  wound-surface  heals  more  rapidly  than  the  ideal  curve 
would  indicate ;  in  this  event  secondary  ulcers  develop  which 
bring  the  curve  back  to  normal.  This  is  the  type  which  is 
represented  by  our  larger  diagram. 

This  application  of  mathematics  to  medicine  is  largely  due 
to  Dr.  Alexis  Carrel  of  the  Rockefeller  Institute  of  Medical 
Research.  He  noted  that  the  larger  the  wound-surface,  the 
more  rapidly  it  healed,  and  that  the  rate  of  healing  seemed  to 
be  proportional  to  the  area.  This  proportionality  constant  is 


LAWS  OF  GROWTH  431 

not  the  same  for  all  values  of  the  surface  or  we  would  have 
an  equation  of  the  form, 

$4»&«"»i 

in  which  Si  is  the  area  at  the  time  that  the  wound  is  rendered 
sterile  and  observations  to  be  plotted  really  begin. 

The  actual  formulas,  as  developed  by  Dr.  P.  Lecomte  du 
Nolly  of  Base  Hospital  21,  Compiegne,  France,  are 

(1)  ;=Si-&> 


giving  the  characteristic  constant  of  the  wound. 

Si  is  the  measure  of  the  area,  first  observation;  S2  is  a 
second  measurement  taken  after  4  days. 

(2)  S.  =  S.-i[l-i'(±  +  V4~n)], 

wherein  Sn  is  the  area  after  4  n  days  ;  similarly,  Sn_i  is  the  area 
after  4(n  —  1)  days,  etc.  ;  each  ordinate  is  obtained  from  the 
preceding  ;  i  is  the  constant  as  determined  above. 

Recent  experiments  by  Dr.  du  Nolly  show  that  there  is  a 
normal  value  of  i  dependent  upon  the  age  of  the  individual 
and  the  size  of  the  wound,  and  that  the  individual  index  as 
determined  by  two  observations  will  doubtless  reveal  facts 
concerning  the  general  health  of  the  individual. 

The  data  given  are  taken  from  the  Journal  of  Experimental 
Medicine,  reprints  kindly  furnished  by  Major  George  A. 
Stewart  of  the  Rockefeller  Institute.  The  diagrams  are 
reproduced  from  the  issue  of  Feb.  1,  1918,  pp.  171  and  172, 
article  by  Dr.  T.  Tuffier  and  R.  Desmarres,  Auxiliary  Hospital 
75,  Paris. 

6.  Damped  vibrations.  —  The  combination  by  multiplication 
of  ordinates  in  the  two  functions,  y  ==  e~*i'  and  y  =.sin  k<£,  which 
we  have  seen  to  be  fundamental  in  the  mathematical  interpre- 
tation of  many  phenomena  of  nature,  gives  a  formula  which 
also  has  wide  application. 

The  formula 

y  =  e~*»*  sin  k<£, 


432 


UNIFIED  MATHEMATICS 


expresses  the  law  by  which  the  decrease  in  intensity  of  the 
vibrations  defined  by  y  =  sin  k.2t  may  be  determined,  under 
certain  conditions. 


Damped  vibration  curve  by  multiplication  of  ordinates 

Thus  a  pendulum  swinging  in  the  air,  when  the  friction  is 
proportional  to  the  velocity,  has  this  form  of  equation  as  the 
equation  of  motion.  We  have  indicated  on  our  diagram  the 

curves 


/ 
y  =  e~S 

and  y  =  —  e~( 
and    the   damped   vi- 
bration curve 
y  =  e~l  sin  2  TT£. 
The  student  should 
check  the  values,  re- 
drawing all  the  curves 
on  double  the  scale  of 
the  illustration  in  the 
text. 

A  beautiful  damped  vibration  curve  is  obtained  by  the 
discharge  of  an  electrical  condenser.  In  our  illustration  the 
equation 

v  =  e~*1  sin  2  irt 


Damped  vibration  produced  electrically  by  the 
discharge  of  a  condenser 


LAWS  OF  GROWTH  433 

represents  quite  closely  the  curve,  using  the  maximum  ordi- 
nate  as  unity,  i.e.  when  t  =  0,  and  on  the  horizontal  axis  the 
measure  of  the  time  in  seconds  of  one  complete  vibration  is 
taken  as  unity ;  in  the  electrical  occurrence  represented  by 
this  photograph,  and  the  corresponding  light  phenomenon 
which  produced  the  photograph,  the  action  took  place  in  about 
•-^j  of  one  second,  and  -^^  of  one  second  is  approximately  the 
time  of  one  vibration  on  this  curve. 


PROBLEMS 

1.  Plot  the  following  curves,  in  the  order  given  : 

y  =  sin  2  irtr, 

taking  two  half-inches  to  represent  t  =  1  on  the  horizontal 
axis  and  6.2  half-inches  to  represent  unity  on  the  vertical  axis  ; 
use  the  graphical  method. 

y  =  «~», 

taking  6.2  half-inches  to  represent  unity  on  the  vertical  axis, 
and  two  half-inches  to  represent  one  second  on  the  horizontal 

axls-  y  =  e~'*  sin  2irf, 

by  multiplication  of  ordinates. 

Note  that  by  taking  five  half-inches  to  represent  unity  on  the  vertical 
axis  each  half  -inch  represents  .2  and  each  twentieth  of  an  inch  represents 
.02.  These  facts  are  to  be  used  when  you  multiply  ordinates  to  obtain  the 
third  of  these  curves.  For  the  values  of  the  powers  of  e  consult  the  table 
at  the  back  of  the  book. 

2.  Given  that  an  automobile  wheel  which  is  revolving  freely 
at  the  rate  of  400  revolutions  per  minute  is  allowed  to  come  to 
rest  by  the  action  of  the  friction  and  air  resistance  ;  assuming 
that  the  subsequent  velocities  per  minute  are  given  at  the  end 
of  t  minutes  by  the  equation 


to  determine   these   velocities,  plot   the  graph  of   the  func- 
tion.    At   what    time   will    the    number    of    revolutions    be 


434  UNIFIED  MATHEMATICS 

reduced  to  approximately  200  per  minute?  to  100?  to  50? 
to  3?  It  may  be  assumed  that  at  3  revolutions  per  minute 
the  law  no  longer  holds  and  that  the  wheel  will  stop  at  about 
that  time. 

3.  Given   that    the   horizontal    displacement   of  a   second 
pendulum  is  4  inches,  and  that  the  horizontal  displacement  is. 
given  by  the  equation 

x  =  4  cos  2  irt, 

and  that  the  amplitudes  are  decreased  according  to  the  "  law 
of  organic  decay,"  the  position  being  given  at  any  instant  by 
the  equation  *  =  4  e^i  cos  2  ^. 

Find  the  displacement  of  the  pendulum  after  10  seconds  ;  after 
100  seconds  ;  after  one  hour.  When  does  the  pendulum  have 
a  displacement  of  only  1  inch  ?  of  ^  inch  ?  of  y1^  inch  ? 

This  type   of   retardation   is   found   when   the   friction   is 
proportional  to  the  velocity. 

4.  Given  that  a  fly-wheel  revolving  freely  with  a  velocity  of 
500  revolutions  per  minute  is  allowed  to  come  to  rest.     If 
the  velocity  at  the  end  of  t  seconds  is  given  by  the  equation 


find  the  velocity  at  the  end  of  10  seconds  ;  at  the  end  of  100 
seconds  ;  at  the  end  of  30  seconds.  When  will  the  velocity  be 
reduced  to  1  revolution  per  minute  ? 

With   a   heavy   oil   as   lubricator   heavy   fly-wheels   follow 
approximately  this  law. 


CHAPTER   XXVIII 

POLAR  COORDINATES 

(See  Section  3,  Chapter  VII) 

1.  Uses. — For  many  purposes  the  representation  of  func- 
tions by  the  system  of  polar  coordinates  is  desirable.     Thus, 
effective  pressure  on  the  crank  head  by  the  piston  head  varies 
for    every    angle.     It    is    convenient    to  give    this   pressure- 
diagram  in  polar  coordinates  ;    on  every  radius  is    plotted  a 
length  representing  graphically  the  effective  turning  pressure 
on  the  crank  for  that  angle. 

2.  Plotting  in  polar  coordinates.  —  The  coordinates  of  points 
which    satisfy   an   equation   given   in   polar   coordinates   are 
obtained  precisely  as  in  rectangular  coordinates.     An  equation 
in   polar   coordinates   involves   r  and   0,   radius   vector   and 
vectorial  angle  ;  by  substituting  in  the  given  equation  particular 
values  of  one  of  the  variables  and  solving  for  the  correspond- 
ing values  of  the  other  points  on  the  curve  are  obtained. 


Illustrative  problem.  —  Plot  the  curve 

3.42 

5.00  r  =  10  sin  20. 

6.43 

7.66          Note    that    when    0  =  10°,    r  =  10   sin  20°  =  3.42,    which 


35° 
40° 
45° 
60° 


9.40      ing  how  the  second  loop  and  other  loops  are  obtained,  by 
9.85      giving  to  0  values  increasing  by  5°  intervals  up  to  360°.     Note 
10         that  no  further  computation  is  needed.     Follow  the  progress 
9.85      of  the  curve  on  the  diagram  given  on  the  next  page. 
435 


436 


UNIFIED  MATHEMATICS 

i«         4 
v<& 


The  formulas  for  transformation  from  rectangular   coordi- 
nates to  polar  coordinates  should  be  noted : 

x  =  r  cos  6, 
y  =  r  sin  6. 

The  formulas  for  transformation  from  polar  to  rectangular 

coordinates  are  : 

r= 


•^ 


^ 


j?/ 


:R 


=  arctan     , 
x 


and      sin  0  = 


or         cos  0  = 


POLAR  COORDINATES  437 

PROBLEMS 

1.  Plot  the  curve  r  =  10  sin  6.     Prove  that  this  is  a  circle. 
Take  any  point  on  the  circle,  as  (r,  0),  and  show  that  the  co- 
ordinates satisfy  the  equation. 

2.  Plot  the  following  curves  : 

(a)  r  sin  6  =  5. 
(6)  r  cos  0  =  10. 

(c)  r  sin  (0-30°)  =10. 

(d)  r  =  10  sin  30. 

(e)  r  =  10  -  10  cos  0. 
(/)  r  =  6. 

3.  Plot   the  curve   r  =  2  a  tan  0  sec  0  ;  transform  to  rec- 
tangular coordinates.     This  curve  is  a  "  cissoid  "  and  can  be 
used  in  the  "duplication.  of  the  cube"  problem. 

4.  Plot  the  curve  r  =  10  sec  0  -j-  5.     This  is  a  "  conchoid  of 
Nicomedes,"  and   can   be  used   to  effect   the  solution  of  the 
problem  to  trisect  any  angle. 

5.  Plot  the  polar  diagram  of  effective  pressures   on   the 
crank  for  different  angles  of  0  ;  use  the  data  of  problem  2  in 
the  problems  given  under  piston-rod  motion.* 

6.  Plot  the  curve  r  =  10  —  10  cos  0.     This  curve  is  called 
a  "  cardioid  "  because  of  its  shape. 

7.  Plot    the   curve   r  =  10  —  5    cos  0.      This   is    called    a 
"  limaqon  of  Pascal." 

8.  Plot  r  =  10  —  20  cos  0,  another  type  of  limaQon. 

9.  Show  that  the  polar  equation  of  any  conic  is 

2m 

f  —  _ 

1  —  e  cos  0  ' 
wherein  2  m  is  one  half  of  the  right  focal  chord. 

10.   Plot  the  parabola 

r=       10 


1  —  cos 


438  UNIFIED  MATHEMATICS 

11.    Plot  the  hyperbola 

r=         10 


1  -  2  cos  B 

For  what  values  of  0  is  r  infinite  in  value  ?  What  directions 
do  these  values  give  ?  Are  these  lines  from  the  origin  then 
the  asymptotes  ? 

12.  Plot  the  spiral  of  Archimedes  given  by 

r  =  10  6. 

13.  Plot  the  hyperbolic  spiral  given  by 


14.  Plot  r=sin20. 

15.  Plot  r  =  sin  0  +  sin  2  0,  and  compare  the  polar  with  the 
Cartesian  (x,  y)  representation. 


CHAPTER   XXIX 
COMPLEX    NUMBERS 

1.  Object.  —  In  the  study  of  the  number  field,  indicated  in 
our  first  chapter,  we  found  that  in  the  extraction  of  square  roots 
we  were  limited  to  positive  numbers.  Again  in  solving  quad- 
ratic equations,  and  in  the  discussion  of  the  roots  of  algebraic 
equations,  we  found  that  no  number  of  the  kind  we  had  con- 
sidered could  occur  as  the  even  root  of  a  negative  quantity. 
We  can  extend  the  number  field,  removing  the  limitation  that 
square  roots  and  even  roots  must  be  taken  of  positive  quan- 
tities only,  by  creating  another  class  of  numbers,  complex 
numbers.  These  numbers,  after  the  fundamental  operations 
with  them  have  been  properly  defined,  apply  to  our  algebraic 
equations,  x  and  the  constants  being  complex  numbers.  In 
the  extended  number  field  it  is  possible  to  prove  that  every 
rational  integral  equation  has  a  root  and  that  such  an  equation 
of  the  nth  degree  has  n  roots. 


2.  Complex  numbers.  —  We  define  V—  1,  designated  by  ?',  as 
a  number  which,  multiplied  by  itself,  equals  —  1 ;  this  re- 
quires, then,  an  extension  of  the  meaning  of  multiplication 
and  a  reexamination  of  the  fundamental  processes  as  applied 
to  the  old  numbers  with  this  newly  found  number  and  other 
new  numbers  which  follow  directly  from  it.  This  discussion 
is  given  graphically  in  the  next  section.  The  square  root 
of  any  other  negative  number,  —a,  is  regarded  as  VaV  —  1, 
or  Va  •  i.  Such  a  number,  e.g.  V— 7,  is  called  a  pure  imag- 
inary. To  add  a  pure  imaginary  to  a  real  number  both  must 
be  written  and  the  combination  is  called  a  complex  number. 

439 


440 


UNIFIED  MATHEMATICS 


Thus,  x  +  yi  and  a  +  bi,  or  —  3  + 2V—  1  and  V5  —  V3i. 
are  complex  numbers. 

Addition  and  multiplication  are  explained  graphically  in 
sections  6  and  7  below. 


3.  Graphical  representation.  —  Our  real  numbers  can  all  be 
conceived  graphically,  as  well  as  analytically,  as  derived  from 
the  unit  1.  Thus,  integers  are  obtained  by  the  repetition  of 
the  unit ;  fractions  are  obtained  by 
the  subdivision  of  the  unit ;  and  nega- 
tive numbers  are  obtained  from  the 
negative  unit,  which  in  turn  is  ob- 
tained by  reversing  the  direction  of 
the  positive  unit.  Analytically  the 
imaginary  unit  repeated  as  a  factor 
gives  —  1 ;  graphically  then  we  would 
desire  an  operation  which  repeated 
gives  a  reversal  of  direction.  How 
is  the  reversing  from  -f  1  to  —  1 
effected?  Evidently  by  turning  the 
an  angle  of  180°  or  -180°.  The 

the 


The    imaginary    unit    ob- 
•  tained  graphically 


positive   unit   through 

be   regarded   then    as   represented   by 


V  —  1,  or  i,  can 
middle  position  of  this  rotating  unit,  and  the  upper  position  is 
regarded  as  +  i  and  the  lower  as  —  i.  This  vertical  line  is 
taken  as  the  axis  of  pure  imaginaries.  Thus,  V  —  4,  or  2  i, 
is  represented  two  units  up  on  this  axis  and  —  V—  2  is  repre- 
sented V2  units  down  on  this  axis. 

A  complex  number,  x  +  yi,  may  now  be  uniquely  represented 
by  the  point  (or,  y]  in  the  complex  plane,  in  which  the  y-axis 
coincides  with  the  vertical  axis  of  pure  imaginaries. 

The  fundamentally  important  facts  concerning  these  num- 
bers are : 

1.  Complex  numbers  are  combined  according  to  the  laws  of  the 
real  numbers  (which  we  have  discussed  in  the  first  chapter), 
noting  that  i2  =  —  1. 


COMPLEX  NUMBERS 


441 


2.  The  combination  of  any  two  or  more  complex  numbers,  by 
the  operations  of  addition,  subtraction,  multiplication,  division 
(except  by  zero),  involution,  and  evolution  (with  certain  excep- 
tions), ALWAYS  produces  a  complex  number. 


mASmm 


Representation  of  three  complex  numbers 

Two  complex  numbers  of  the  form  x  +  yi  and  x  —  yi,  sym- 
metrically pla,ced  with  respect  to  the  axis  of  reals,  are  called 
conjugate  complex  numbers.  Their  sum  and  their  product  are 
real  numbers. 

4.  Complex  roots  in  pairs.  —  In  any  rational  integral  algebraic 
equation  with  real  coefficients,  if  a  -f  bi  is  a  root  of  theA  equa- 
tion, then  a  —  bi  is  also  a  root  of  the  equation.  The  proof 
depends  upon  the  fact  that  when  a  4-  bi  is  substituted  in 

a^cn  +  a^x"'1  -f  a?xn~z  -f  •  ••  an, 

the  resulting  expression  is  of  the  form  P  -f  Qi,  in  which  P  and 
Q  being  real  numbers,  P  involves  powers  of  a  and  the  even 
powers  of  bi,  and  Q  is  obtained  from  expressions  involving 
odd  powers  of  bi.  Now  if 


then  P  =  0  and  Q  =  0  ;  otherwise   you   have   a   real   number 
equal  to  a  pure  imaginary.     Substituting  a  —  bi  for  x  in 


442 


UNIFIED  MATHEMATICS 


changes  the  signs  of  the  terms  involving  the  odd  powers  of  i, 
and  does  not  change  the  sign  of  the  even  powers.  Hence 
a  —  bi  substituted  gives 


but  P=0  and  Q  =  0,  hence  P—  Qi  =  Q.  Therefore  a  —  bi  is 
also  a  root  of  the  equation  if  a  +  bi  is  a  root.  Complex  roots 
go  in  pairs. 


Illustrative  problem. — 1.    Find  the  product  of  2-fV— 3  by 
3  —  V  —  2  and  put  the  product  in  the  x  -f  yi  form. 


Graphical  representation  of  the  product  of  2  +  V3  i  by  3  — 

(2  +  V3  t)(3  -  \/2  i)  =  6  +  3  v'3"i  -  2  \/2  i  -  \/6  i2, 
but  t2  =  —  1  ,  giving  as  product, 

6+V6+(3V3  -2\/2)i.    Ans. 
6+V6=  a;  3V3-2V2  =  6.  • 


2.   Divide     3  +  V—  3    by    5  —  2V  —  3   and     express     the 
quotient  in  x  +  yi  form. 


3  +  V3t 


t)(5  +  2V3t)   _  15+ 


_2V^3      5-2V3i      (5  -  2  V3  t)(5  +  2V3  i) 
_  9  +  llv'3i  _  _9_      llv/3t 
37          ~  37          37 


25-12? 


3.   Factor  xz  +  y2  into  complex  factors,  linear  in  x  and  y. 
x*  +  yz  =  (x  +  iy)  (x  —  iy). 


COMPLEX  NUMBERS  443 


PROBLEMS 
1.   Write  the  conjugate  complex  numbers  : 

a.  3  +  V^.  cl   !  +  «'• 

b.  -4-21  e. 


2.  Rationalize  the  denominator  in  the  following  expressions 
by  using  the  conjugate  complex  number  as  multiplier,  reduc- 
ing the  quotient  obtained  in  this  way  to  the  form  a  +  bi. 


a.    5~V~2.  a.    J- 


b. 


3  +  V-2 

5  V5 


-  4  -  2 1 
c.  — •  £ 


3 

3.   Write  the  following  expressions  in  the  form  a  +  bi: 
a.  t-  +  i*  +  t»+i«.  2  2 

€'       n    .    n    •> 


6.    t-5  .(_  3  (•  -j.  3  p  +  4  i8.  3  +  2*3-2  i 

c.    t^+i20.  2  2 

d. 


4.  Locate  the  points  represented  by  the  complex  numbers 
in  problem  1. 

5.  Square    -    -  +  —  -z;  square  -  -  —  ^-f.     These  are  roots 

L        £  2i        £ 

of  a^-  1  =  0.     Multiply  _l  +  ^»  by-  \-^j>-i.    What  is 

22  22 

the  cube  of  ---f^t? 

1  —  t  l  +  i 

6.  Square  -      -  and  -    —  .     Give  an  equation  with  real 

V2  V2 

coefficients    which    these    numbers    satisfy.     What   are    the 
square  roots  of  i? 


444 


UNIFIED  MATHEMATICS 


5.  Vectors.  Polar  representation  of  complex  numbers.  —  The 
complex  number  x  +  yi  may  be  regarded  either  as  determined 
by  the  point  P(x,  y)  or  by  the  vector  OP,  which  by  its  length 
and  direction  determines  the  position  of  P.  The  angle  which 
OP  as  a  ray  makes  with  the  sc-axis  is  called  0,  the  amplitude  or 
angle,  and  the  length  OP  is  called  r,  the  modulus  of  the  com- 
plex number.  In  other  words,  the  polar  coordinates  of  P  are 

faff); 

r  =  Vx2  -(-  y2, 

cos  0  =  - ,  and  sin  6  =  ^. 
r  r 

The  complex  number  may  be  writ- 
ten in  the  form 

r(cos  6  +  i  sin  0), 
which  is  termed  the  polar  form. 


t  0:rr 


i 


Modulus  and  amplitude  of  . 

a  complex  number  Tlie    modulus,   r   or   V»2  +  y2,   is   a 

positive    number    representing    the 

length  of  the  vector  or  the  distance  of  the  point  x  +  iy  from 
the  origin.  This  modulus  is  sometimes  called  the  stretching 
factor  or  the  tensor;  see  section  7. 


6.  Addition  of  vectors.  —  "\Vhen  the  complex  number  is  repre- 
sented by  a  vector,  the  sum  of  two  complex  numbers  will  be 
represented  by  the  diagonal  of  the  parallelogram  formed  by 
the  two  vectors ;  see  Chapter  IX,  section  2.  The  student 
should  verify  the  fact  by  a  diagram. 


7.  Product  of  complex  numbers.  —  Given  two  complex  num- 
bers, either  in  polar  form  or  in  rectangular  form,  the  product 
of  the  two  numbers  is  also  a  complex  number ;  further  the 
modulus  of  the  product  is  the  product  of  the  moduli,  and  the 
amplitude  or  angle  of  the  product  is  the  sum  of  the  amplitudes 
of  the  factors. 


COMPLEX  NUMBERS  445 

Let 

r^cos  61  +  i  sin  0^  ;  xv  +  yj,  mod  Va^2  +  y^,  ampl  0X  =  tan"1  ^ 
and 

r2(cos  02  4- 1  sin  02) ;   a^  +  */2*>  mod  Va^2  +  2/22>  ampl  02  =  tan"1  — 
be  two  complex  numbers.     Their  product  is 
7'ir2[cos  $i  cos  $2  —  sin  $1  gin  $2  4-  *'(sin  $1  cos  ^2  4-  cos  $1  sijl  $2)] 
#1^2  —  */i#2  4-  *  (^12/2  4-  #22/i)>   m°d  Vajj2^2  4-  2/i2?/22  4-  ^i2^2  +  x^y^ 
ampl  03  =  tan    — s— '•  • — . 

The  polar  product  may  be  written 

rtr2[cos  (0t  4-  02)  +  i  sin  (0X  +  02)], 

showing  that  the  product  of  the  moduli  i\  and  r2  is  the  modulus 
rtr2  of  the  product  and  the  amplitude  is  0X  +  02,  the  sum  of  the 
amplitudes.  It  is  left  as  an  exercise  for  the  student  to  show 
that  the  analytical  expressions  for  modulus  and  amplitude 
establish  the  same  facts. 

When  any  complex  number  is  used  as  a  multiplier,  the 
modulus  of  the  product  is  the  modulus  of  the  multiplicand 
stretched  in  the  ratio  of  the  modulus  of  the  multiplier  to 
unity.  For  this  reason  the  modulus  is  sometimes  termed  the 
stretching  factor. 

8.  De  Moivre's  theorem.  —  Evidently,  if  0t  and  02  are  set 
equal  to  0,  our  product  formula  may  be  written  : 

(1)  [r(cos  0  4-  i  sin  0)]2  =  r2(cos  2  0  4  t  sin  2  0). 
Evidently  by  mathematical  induction,  by  simple  introduction 
of   one  further  factor  ?-(cos  0  + 1  sin  0)   at  a  time,  it  can  be 
shown  that 

(2)  [r(cos  0  4-  i  sin  0)]n  =  r"(cos  nO  +  i  sin  w0). 

This  theorem,  which  holds  for  all  values  of  n,  is  called  De 
Moivre's  theorem.  We  have  proved  it  only  for  n  an  integer. 
Taking  the  wth  root  of  each  member  of  equation  (2),  we  have 

(rn) "  (cos  ?i0  +  i.  sin  nff) n  =  r(cos  0  4  i  sin  0). 


446  UNIFIED  MATHEMATICS 

Let  r"  =  k  and  nO  =  0',  which,  as  no  limitation  was  imposed  on 
6,  imposes  no  limitation  as  to  value  on  6',  and  we  have 

[k(cos  6'  +  i  sin  0')]"  =  k  "(cos  -  +  i  sin  -Y 
\       n  n  J 

i.e.  our  formula  holds  for  a  fractional  exponent  of  the  form  -. 

n 

By  raising  to  the  mth  power  both  sides,  it  can  be  shown  to 
hold  for  any  fractional  exponent. 

For  ?i  =  — 1,      [r(cos6  +  izwff)~\-1  = 

r(cos  9  +  i  sin  0) 

cos  0  —  i  sin  6 
=  r(cos2  6  +  sin2  6) 
=  r-1(cos  0—i  sin  0), 
whence 

[r(cos  0  +  i  sin  0)]"1  =  r^[cos  (-  0)  +  i  sin  (—  0)], 

which  establishes  the  formula  when  n  =  —  1.  By  raising  both 
sides  to  the  nth  power,  n  any  rational  number,  the  theorem  is 
established  for  all  rational  exponents. 

The  theorem  can  be  established  also  for  irrational  values 
of  n. 

PROBLEMS 

1.   Write  the  following  complex  numbers  and  their  conju- 
gates in  polar  form,  giving  modulus  and  amplitude : 


b     -4-2i  g' 

c.  _3-V2-V^.  h'   23' 

d.  l  +  i._  ?.         1      V3\ 

2.  Show  that   (cos  30°  + 1  sin  30°)2  =  cos  60°  +  i  sin  60°,  by 
multiplication. 

3.  Show  that  (cos  30°  +  z  sin30°)8  =  z,  i.e.  cos  90°  +  i  sin  90°. 

4.  Show  that  -  -  =  cos  ( -  30°)  +  i  sin  ( -  30)°. 

cos  30°  +  i  sin  30° 


COMPLEX  NUMBERS  447 

5.  Show  that  (cos  60°  +  i  sin  60°)*  =  cos  30°  +  i  sin  30°. 

6.  What  is  the  value  of  (cos  30°  +  i  sin  30°)*  by  De  Moivre's 
theorem  ? 

7.  Plot,    using    2    inches    as   1    unit,     cos  15°  +  i  sin  15°, 
cos  30°  +  i  sin  30°,  cos  45°  +  i  sin  45°. 


o/jrvo 

8.  Plot    the    point  B,   cos  --  1-  i  sin  -  ;    connect   by   a 

chord  with  the  point  A,  cos  0°  +  *  sin  0°  ;  take  this  length  as  a 
chord,  successively  seven  times  on  the  unit  circle  about  O.  This 
chord  is  the  side  of  a  regular  inscribed  polygon  of  seven  sides. 

9.  Roots  of  unity.  —  Plotting  the  solutions  of  the  following 
equations  on  the  complex  number  diagram, 

(See  Section  3,  above), 
x  —  1  =  0  gives  one  point,  1  ; 
a?  —  1  =  0  gives  two  points,  1  and  —  1  ; 

a^-l  =  0  gives  three  points,  1,  -  5  +  -5-  *'  ~    >~~^~7'» 

u          m  a          Ji 

s*  —  1  =  0  gives  four  points,  1,  —  1,  i  and  —  i  ; 
of  —  1  =  0  gives  six  points, 


x8  —  1  =  0,  or  (x4  —  l)(x4  +  1)  =  0,   gives   eight   points,  which 
may  be  obtained  by  methods  of  quadratic  equations.     For 

a-*  +  1  =  ^  +  2  x2  +  1  -  2  a?  =  (&  +  I)2  -(V2  «)» 
=  (a2  +  1  -  V2  z)(o;2  +  1  +  V2  x), 


whence,  x2- 
and 


448 


UNIFIED  MATHEMATICS 


a8— 1  =  0  gives  eight  points,  1,  —  1,  L  —  i,  —  4.  — -  t,  - 

V2      V2      V2      V2 

11.  11. 

-| -  i,  and 1. 

V2      V2  V2      V2 

a;5  —  1  =  0   gives   five  points  ;   see  the  solution  obtained  in 
problem  4,  page  97.     The  solutions  are 


V5  +  1 


-  2  y5  .       V5  +  1  ,  V10-2V5  . 

~      *'          ~         ~         *» 


-1  +  V5 


;.       d 


-1+V5 


Plotting  these  points  on  the  complex  diagram  gives  the  ver- 
tices of  a  regular  pentagon. 

Graphically  representing  these  points  we  have  the  following 
diagrams  : 


Graphical  solutions  of  y?-\  =  0,  s>  -  1  =  0,  x6  -  1  =  0,  x12  -  1  =  0,  x8  - 1  =  0, 

and  x5  -  1  =  0 


COMPLEX  NUMBERS  449 

The  roots  of  any  equation  of  the  form 
xn  —  1  =  0,  n  integral, 

are,  aside  from  +  1  or  —  1,  complex  numbers.  Since  any  real 
number  less  than  1  when  multiplied  by  itself  gives  a  number 
less  than  1  and  since  any  real  number  greater  than  1  multiplied 
by  itself  gives  a  number  greater  than  1,  it  follows  that  the 
roots  other  than  1  or  —  1  are  complex.  Further,  the  modulus 
of  each  of  these  complex  roots  is  a  real  number  which,  taken  n 
times  as  a  factor,  produces  1  ;  hence  the  modulus  of  any  nth 
root  of  unity  is  1.  We  need  then  to  know  only  the  real  part  of 
any  root  of  unity  to  plot  it,  since  the  root  .itself,  having  a 
modulus  1,  lies  on  the  unit  circle,  x2  +  y2  =  1. 

The  nth  roots  of  unity  can  be  obtained  graphically  by  finding 
the  angles  which  repeated  n  times  give  360°  or  integral 
multiples  of  360°. 

Thus  for  the  twelfth  roots  of  unity  these  angles  are  0°,  30°, 
60°,  90°,  120°,  150°,  180°,  210°,  240°,  270°,  300°,  and  330°. 
Writing  the  corresponding  complex  numbers  in  polar  form  we 
have  the  twelve  twelfth  roots  of  unity.  If  we  went  farther, 
taking  360°,  390°,  420°,  •••,  we  would  simply  repeat  values 
already  obtained.  The  twelfth  roots  of  unity  are,  then, 


V3      1  .       1      V3  .  .  1      V3  .  ,  V3     1  . 

—  -  ---  j,  ----  -—  i,  —i.  -\  -----  i,  and  +—  ---  1, 

2222  22  22 

as  on*  the  figure. 

10.  Historical  note.  —  Just  as  negative  numbers  were  gener- 
ally accepted  only  after  a  graphical  scheme  of  representation 
of  these  numbers  was  introduced  by  Descartes,  so  imaginary 
numbers  were  neglected  and  even  rejected  by  mathematicians 
until  a  graphical  system  of  representation  was  found. 

In  1797  a  Norwegian,  Caspar  Wessel,  presented  the  scheme 
of  representation  of  complex  numbers  to  the  Danish  Academy, 
but  public  recognition  of  his  work  is  only  recent.  A  French- 


450  UNIFIED  MATHEMATICS 

man,  J.  R.  Argand,  discovered  the  same  system  independently 
in  1806  and  it  appears  that  the  German  J.  C.  F.  Gauss  in  1831 
again  independently  rediscovered  this  graphical  method.  The 
latter  made  extensive  use  of  the  diagram  and  since  then  these 
numbers  play  a  vital  role  in  the  development  of  algebra. 
Quite  recently  the  practical  importance  of  these  numbers  has 
been  more  generally  recognized  by  physicists  and  engineers. 
Applications  have  been  made  to  problems  in  electricity,  to  the 
steam  turbine  by  Steinmetz,  and  to  numerous  other  vector 
problems. 

The  term  imaginary  is  a  misnomer,  as  our  development 
shows.  So  far  as  actuality  is  concerned,  3  +  V—  3  exists  as  a 
number  quite  as  much  as  3  or  V3  ;  all  numbers  are  the  product 
of  intelligence  reacting  on  the  experiences  of  life,  and  in 
this  sense  all  numbers  are  imaginary,  the  product  of  the  imagi- 
nation. 

1 1 .  Mathematical  unity.  —  The  complex  numbers  are  fittingly 
chosen  to  conclude  our  treatment  of  plane  analytic  geometry, 
elementary  algebra,  and  elementary  trigonometry  since,  as  the 
observant  student  will  have  noticed,  we  have  here  involved 
the  fundamental  principles  of  these  subjects  as  well  as  theo- 
rems of  plane  geometry. 

We  might  note  that  while  regular  polygons  of  seven  and 
nine  sides  cannot  be  constructed  with  ruler  and  compass,  since 
the  solutions  of  these  equations  lead  to  cubics  which  cannot  be 
solved  in  terms  of  quadratic  irrationalities,  there  are  other 
polygons  having  a  prime  number  of  sides  which  can  be  so  con- 
structed. Gauss,  when  only  19  years  old,  showed  that  the 
polygon  of  17  sides,  and,  in  general,  the  polygon  of  sides 
22"  -f  1  in  number,  when  this  number  is  prime,  can  be  con- 
structed with  ruler  and  compass.  The  corresponding  alge- 
braic fact  is  that  xm  —  1  =  0,  when  m  is  a  prime  number  equal 
to  22"  + 1,  is  solvable  by  the  methods  of  quadratics,  and  the 
roots  can  be  expressed  in  functions  involving  only  square 
roots. 


COMPLEX  NUMBERS  451 

PROBLEMS 

1.  Solve  ce3  —  1  =  0  and  plot  the  points  on  the  polar  diagram. 

2.  Solve  similarly  y?  +  1  =  0  and  plot. 

3.  Solve  a;5  4- 1  =  0.     Using  the  results  given  for  of  —  1  =  0, 
write  all  the  solutions  of  cc10  —  1  =  0. 

4.  Derive   the   formulas   for   cos  2  9  and  sin  2  6,  using  De 
Moivre's  theorem. 

HINT,  cos  2  0  +  i  sin  2  0  =  (cos  6  +  i  sin  0)2.  Square  the  right-hand 
member  and  then  equate  cos  2  6  to  the  real  part  and  put  sin  2  0  equal  to 
the  coefficient  of  i. 

5.  Derive  formulas  for  cos  30  and  sin  3d  by  the  process 
of  problem  4. 

6.  Show  on  the  diagram  how  to  obtain  the  twelfth  roots  of 
unity.     What  equation  do  these  numbers  satisfy  ? 

7.  Show  geometrically  and  algebraically  how  you  can  obtain 
the  solutions  of  the  equation 

x*  -  I  =  0       . 

from  the  complex  diagram.  Use  also  the  formulas  for  sin  ^  6 
and  cos \Q  to  obtain  sin  15°  and  cos  15°  from  sin 30°  and 
cos  30°. 

8.  Solve   y?  —  1  =  0,   and  plot   the  points  on  the  complex 
diagram. 

9.  Solve  a;16  —  1  =  0.     What  angles  are  involved  ? 


CHAPTER   XXX 
SOLID  ANALYTIC  GEOMETRY:  POINTS   AND   LINES 

1.  The  third  and  fourth  dimensions.  —  We  have  found  that  on 
a  line  the  position  of  any  point  may  be  given  by  a  single 
number,  x,  which  locates  the  point  with  reference  to  one  fixed 
point  on  the  line.  The  single  number,  commonly  x,  represents 
distance,  in  terms  of  some  unit  of  length,  from  the  point  of 
reference,  and  direction  by  means  of  a  +  or  —  sign.  In  a 
plane  the  position  of  any  point  may  be  given  by  a  pair  of 
numbers  which  locate  the  point  with  reference  to  two  fixed 
lines  in  the  plane.  The  two  numbers,  x  and  y  commonly, 
represent  the  distances  in  determined  order,  in  terms  of  some 
unit  of  length,  from  each  of  the  two  given  lines  of  reference, 
and  direction  as  before.  By  analogy,  continuing  with  the 
proper  changes,  it  is  obvious  that  in  space  the  position  of  any 
point  may  be  given  by  a  set  of  three  numbers  which  locate  the 
point  with  reference  to  three  fixed  planes  in  space.  The  three 
numbers,  x,  y,  and  z  commonly,  represent  the  distances  in 
determined  order,  in  terms  of  some  unit  of  length,  from  the 
three  given  planes  of  reference,  and  the  direction  in  each  case 
is  determined  by  the  algebraic  sign  of  the  number.  If  the 
analogy  could  be  continued  we  could  state  that  the  position 
of  any  point  in  a  four-dimensional  space  would  probably  be 
given  by  a  set  of  four  numbers  which  locate  the  point  with 
reference  to  four  given  "  three-dimensional "  spaces.  The  four 
numbers,  x,  y,  z,  and  w  commonly,  would  then  represent  the 
"distances"  in  determined  order,  in  terms  of  some  unit  of 
length,  from,  each  of  the  spaces  of  reference.  Without  a  pre- 
cise definition  of  what  we  mean  by  "  distance  "  of  a  point  in 

452 


SOLID  GEOMETRY:  POINTS  AND  LINES        453 


"  four-dimensional  space  "  from  a  "  three-dimensional  space  " 
these  analogies  must  be  regarded  as  purely  fanciful,  and  devoid 
of  physical  significance. 


LOCATION   OF   A   POINT 


Upon  or  in  a 

With  reference  to 

£y  means  of 

line, 

one  point, 

one  variable, 

one  dimensional. 

zero  dimensional. 

X. 

plane, 

two  lines, 

two  variables, 

two  dimensional 

one  dimensional. 

(»,  y). 

space  (ordinary), 

three  planes, 

three  variables, 

three  dimensional. 

two  dimensional. 

.  O,  y,  z). 

hyperspace, 

four  three-spaces, 

four  variables, 

four  dimensional. 

three  dimensional. 

(x,  y,  z,  w>). 

n-space, 

n(n  —  1)  -spaces, 

n  variables, 

n  dimensional. 

(»—  1)  dimensional. 

(Xi,  X2,  X3,  "-Xn). 

2.  Space  coordinates.  —  The  position  of  a  point  in  ordinary 
space  is  determined  by  location  with  respect  to  three  inter- 
secting planes,  called  the  coordi- 
nate planes.  Just  as  our  lines  of 
reference  were  chosen  perpen- 
dicular to  each  other,  for  conven- 
ience, in  plane  analytics,  so  here 
the  planes  of  reference  are  taken 
mutually  perpendicular,  like  the 
three  sides  of  a  box  or  like  the 
front  wall,  the  floor,  and  the  left- 
hand  wall  of  a  room.  The  three 
lines  of  intersection  of  these 
planes  with  each  other  in  pairs 
are  called  the  axes  of  coordinates, 


Axes  in  space 


designated  as  x-axis,  y-axis,  and  z-axis;  the  three  planes  are 
named  xy-,  xz-,  and  yz-planes  respectively ;  the  point  common 
to  the  three  planes  and  to  the  axes  is  called  the  origin. 


454 


UNIFIED  MATHEMATICS 


•-& 


The  numbers  x,  y,  and  z  represent  respectively  distances 
from  the  yz-,  xz-,  and  ay-planes ;  direction  is  determined  as 
indicated  by  the  arrowheads  upon  the  diagram.  No  general 
agreement  has  been  reached  as  to  which  axis  to  use  as  the 
vertical  axis.  The  system  shown  is  called  a  right-handed 

system,  since  the  90°  rotation 
of  the  positive  ray  of  the 
a-axis  into  the  positive  y  ray 
advances  a  right-handed  screw 
along  the  z-axis,  and  similarly 
with  the  other  axes,  by  cyclical 
interchange  in  x,  y,  z  order. 
The  positive  directions  of 
these  three  axes  can  be  repre- 
sented by  the  thumb,  first 
finger,  and  second  finger  of 
the  right  hand. 

To  determine  the  coordi- 
nates of  any  point  P  in  space, 
planes  are  drawn  or  conceived 
through  the  point  parallel  to  the  coordinate  planes ;  the  dis- 
tances OL,  OM,  and  ON  cut  off  on  the  axes  are  given  with 
proper  sign  as  the  coordinates  (x,  y,  z)  of  the  point  P. 

Space  is  divided  by  the  coordinate  planes  into  eight  divi- 
sions, called  octants.  The  signs  of  the  coordinates  of  any  point 
within  an  octant  are  given  in  xyz  order  to  distinguish  the 
octants.  Thus  the  +  —  —  octant  is  at  the  right,  below,  and 
back. 

To  every  point  in  space  corresponds  one  set  of  coordinates 
and  -only  one,  and,  conversely,  to  every  set  of  three  numbers 
corresponds  one  and  only  one  point  in  space.  When  a  point 
is  given  by  its  coordinates,  the  position  is  determined  on  the 
diagram  by  passing  a  plane  through  the  a/-axis  at  the  x  of  the 
point,  parallel  to  the  t/z-plane ;  on  the  intersection  of  this 
plane  with  the  ccy-plane  indicate  the  y  coordinate.  The  third 
coordinate  must  be  represented  in  perspective,  and  the  direc- 


O(0,  0,  0);   L(3.5,  0,  0);  M (0,3,0); 

N(0,  0,  2) ;  CK3.5,  0,  2) ;  R(3.5,  8,  0) ; 

5^0,3,2);  />(3.5,  3,  2) 


SOLID  GEOMETRY:  POINTS  AND  LINES       455 

tion  and  length  of  these  units  in  perspective  are  taken  paral- 
lel and  equal  to  the  units  represented  on  the  third  axis. 


Drawing  by  L.  Makielskl. 
An  artist's  conception  of  a  rectangular  solid  in  space 

3.  Fundamental  propositions  of  solid  geometry.  —  The  follow- 
ing propositions  of  solid  geometry  have  constant  application 
in  our  further  work.  The  student  would  do  well  to  review 
these  propositions  in  any  elementary  work  on  solid  geometry 
and  further  to  verify  the  reasonableness  of  these  propositions 
on  our  figures. 

a.    Two  planes  intersect  in  a  straight  line. 

6.  If  a  line  is  perpendicular  to  each  of  two  intersecting 
lines,  it  is  perpendicular  to  the  plane  of  the  two  lines,  i.e.  it 
is  perpendicular  to  every  line  in  the  plane  of  the  two  given 
lines. 

Thus,  PQ  on  the  diagram  below  is  perpendicular  to  QL,  to 
QM,  to  QD,  and  to  every  line  in  the  a;2-plane  which  passes 
through  Q.  A  line  in  the  #2-plane  which  does  not  pass  through 
Q  does  not  intersect  PQ,  but  the  angle  which  it  makes  with 
PQ  is  defined  as  the  angle  which  any  parallel  to  it  which  does 
intersect  PQ  makes  with  PQ.  This  will  then  be  a  right  angle. 

c.  The  angles  between  two  pairs  of  parallel  lines  are  equal 
or  supplementary. 

d.  If  two  planes  are  perpendicular  to  a  third,  their  inter- 
section line  is  perpendicular  to  the  third. 


456 


UNIFIED   MATHEMATICS 


e.  The  dihedral  angle  between  two  planes  is  measured  by 
the  plane  angle  formed  by  two  lines,  one  in  each  plane,  both' 
perpendicular  to  the  edge  of  the  dihedral  angle. 

/.    If  a  line  (PQ)  is  perpen- 


The  dihedral   angle,  between  two 

planes,  is  measured  by  the 

angle  between  two  lines 


dicular  to  a  plane  (xz)  and  from  the  foot  of  the  perpendicular 
a  second  perpendicular  (QL  or  QD)  is  drawn  to  any  line 
(OX  or  NL]  in  the  plane  (zx),  then  the  line  connecting  any 
point  (P)  on  the  first  perpendicular  to  the  intersection  point 
(L  or  Z>)  is  perpendicular  to  the  line  (QX  or  NL)  in  the  plane. 

NOTE.  —  We  will  refer  to  these  propositions  as  3  a,  36,  3  c,  3d,  3e, 

and  3/. 


4.  Vectorial  repre- 
sentation. —  For 
some  purposes  it  is 
convenient  to  think 
of  three  numbers  (x, 
y,  z)  as  representing 
the  vector  from  the 
origin  to  the  point. 
The  length  OP  of 
the  vector  is  called 
r,  and  the  vectorial 


Point  P(x,  y,  z)  or  P(r, 


SOLID  GEOMETRY:  POINTS  AND   LINES        457 

angles  which  this  vector  makes  with  the  x-,  y-,  and  z-axes  are 
termed  direction  angles  and  are  represented  by  the  letters, 
«,  ft,  and  y,  respectively. 

By  proposition  3  /,  the  triangles  PLO,  PQO,  and  PNO  are 
right  triangles.     Hence 

x  =  r  cos  a         =  r  cos          z  =  r  cos    . 


Further,  the  square  on  the  diagonal  OP(  =  r)  of  our  rectangu- 
lar prism  is  the  sum  of  OQ2  and  QP2,  and  OQ2  =  ON*  +  NQ>, 
whence 

r2  —  r2  cos2  a  +  r2  cos2  ft  +  y2  cos2  y, 

or  cos2  «  +  cos2  /3  +  cos2  y  =  1. 

Evidently,  also, 

r2  =  z2  +  y2  +  z2. 

5.    Parametric  equations  of  a  line.  —  The  equations 
x  =  r  cos  a,      y  =  r  cos  /?,      z  =  r  cos  y, 

when  «,  ft,  and  y  are  fixed  and  r  is  a  variable  parameter,  serve 
as  the  equations  of  the  straight  line  OP.  Cos  a,  cos  ft,  and 
cos  y  are  called  direction  cosines  of  this  line. 

Evidently  any  point  E(a,  b,  c)  on  the  line  OP  produced  in 
either  direction  satisfies  this  relationship,  if  r  is  taken  as  the 
distance  from  0  to  E.  If  E  is  on  the  other  side  of  the  origin 
from  P,  then  the  direction  angles  of  the  vector  OE  are  supple- 
mentary to  a,  ft,  and  y  and  have  cosines  opposite  in  sign  to 
cos  a,  cos  ft,  cos  y.  In  this  case  r  is  taken  as  negative  and  it  is 
evident  that  with  this  interpretation  the  coordinates  of  the 
point  E  satisfy  the  given  equations. 

A  line  which  does  not  pass  through  0  has  the  same  direction 
cosines  as  the  line  parallel  to  it  through  0,  positive  directions 
on  both  being  the  same.  If  such  a  line  passes  through 
P(XI,  ?/!,  zx)  in  space,  the  corresponding  parametric  equations 
are 

x  —  Xi  =  r  cos  a,  y  —  y\  =  r  cos  ft,  z  —  zl  =  r  cos  y. 


458 


I M  VIED  MATHEMATICS 


In  plane  analytics  the  corresponding  formulas  for  straight 
lines  are 

x  =  r  cos  a,  y  =  r  cos  /?,  where  cos  /8  =  sin  a 

x  —  Xi  =  r  cos  a,  y  —  yl  =  r  cos  ft. 

The  student  should  check  these  formulas  and  show  their  rela- 
tions to  the  ordinary  equations  used. 

6.    Distance  formulas  and  spheres.  — 


is  the  square  of  the  distance  from  (x,  y,  z)  to  (0,  0,  0). 

a;2  +  yZ  _|_  22  _  fi 

is  an  equation  which  is  satisfied  by  every  point  on  a  sphere. 


m 


5= 


= 


E± 


The  sphere  :  (x  -  h)2  +  (y  -  fc)2  +  (z  -  I)-  =  r- 


r  = 


is  the  distance  between  two  points  (xl9  ylf  z^)  and  (x*,,  y.,,  z2). 


represents  the  square  of  the  distance  from  (x,  y,  2)  to  (h,  k,  I) 
and  hence,          (^  _  7^2  +(y-  fc)2  +  (2  _  Z)2  =  r2 

is  the  equation  of  a  sphere  whose  center  is  (h,  k,  /)  and  whose 
radius  is  r. 


SOLID  GEOMETRY:  POINTS  AND  LINES        459 
7.    Point  of  division. 


,  _ 


<>; 


J/2  -  J/3 


-  Z3 


Step  for  step,  and  letter  for  letter,  the  proof  follows  that 
given  for  the  point  of  division  in  a  plane  ;  the  only  change  is 
that  the  z-  term  is  added. 


Thus,  since 


•»     / 


=    1,  it  follows  that 


=     }  etc. 


PROBLEMS 


1.  "\Vhat  does  the  equation  x  =  3  or  x  —  3  =  0  represent  on 
a  line,  the  or-axis,   i.e.  when  you  are  considering  points  on  a 
line  ?     What  does  this  equation  represent  in  plane  analytics  ? 
What  does  this  equation  represent  in  space  analytics  ? 

2.  What  does  the  equation  x2  =  9  represent  in  one  dimen- 
'sioual  analytics?      How  are   these  two   points  located  with 

reference  to  the  origin  ?      What  does  the  equation  x2  +  y2  =  9 


460  UNIFIED  MATHEMATICS 

represent  in  the  ccy-plane  ?  How  are  the  points,  which  lie  on 
this  locus,  located  with  reference  to  the  origin '.'  What  does 
the  equation  x2  +  y2  +  z2  =  9  probably  represent  in  xyz- 
coordinates  ? 

3.  Where  do  all  lines  lie  which  make  an  angle  of   +30° 
with  the  o?-axis  ?     Where  do  all  lines  lie  which  make  an  angle 
of  +  70°  with  the  ?/-axis  ?     Where  does  a  line  lie  which  has 
the  angle  a  equal  to  30°  and  the  angle  ft  equal  to  70°  ?     Deter- 
mine the  angle  y,  using  the  relation  cos2  «  +  cos2  ft  +  cos2  y  =  1. 
Given  that  a  =  30°  and  ft  =  60°,  what  is  the  value  of  y  ?     If 
«  =  30°,  ft  =  45°,  what  is  the  value  of  y  ? 

4.  Write    the    equations,   in    parametric    form,   of    a    line 
through   the  origin  which  has  the  direction  angles  a  =  30°, 
ft  =  70°,  and  the  third  angle  as  determined  in  the  preceding 
problem.     Write  the  equations  when  these  angles  are  30°,  60°, 
and  90°. 

5.  Write  the  equations  of  lines  parallel   to  the  two  lines 
of   the   preceding   problem  and   passing   through    the    point 
(-2,3,  -7). 

6.  Write  the  equation  of  the  sphere  whose  radius  is  10  and 
whose  center  is  the  point  (2,  —  3,  4).     Find  three  other  points 
on  this  sphere. 

7.  Find  the  coordinates  of  the  points  of  trisection  of  the 
line  joining  A(—2,  3,  -  7)  to  B(2,  -  3,  -  4).     If  the  line  AB 
is  extended  through  B  by  its  own  length,  what  are  the  coordi- 
nates of  the  point  so  determined  ? 

8.  Given  that  the  parametric  equations  of  a  line  are 

x  =  3r,  y  =  —  2r^  z  =  —  or, 

find  10  points  upon  the  line  by  giving  to  r  values  from  —  4  to 
+  5.  Determine  the  direction  cosines  of  this  line.  Determine 
from  your  trigonometric  tables  the  angles  «,  ft,  and  y. 

9.  Given  the  parametric  equations  of  a  line 

x  —  3  =  3r,  y  +  5  =  —  2r,  and  z  —  7  =  —  5  r, 
find  10  points  on  the  locus ;  determine  the  direction  cosines  ' 
and  the  angles  a,  ft,  and  y. 


SOLID  GEOMETRY:  POINTS  AND  LINES       461 

8.  Angle  between  two  lines.  —  We  have  had  occasion  to  note 
that  the  angle  between  two  non-intersecting,  or  skew,  lines  in 
space  is  defined  as  the  angle  be- 
tween two  intersecting  lines  which 
are  respectively  parallel  to  these 
given  lines.  For  convenience 
these  parallels  may  be  taken 
through  0. 

The  angle  0  between  two  lines 
having  direction  cosines  («i,  fa,  yt) 
and  («2,  /8s,  72)  respectively  is  ob- 
tained as  follows : 


Take  any   two   points    Pt  and 


The  angle  between  two  lines  in 
space 


P2,   one   on   each   line,  having   vector    distances    rt    and    r2. 
Evidently 

I\P?=  r?  -f  r22  -  2  ?v-2  cos  0. 
But 


Equating  the  two  values  and  canceling,  noting  that 

ri2  =  Xl*  +  y?  +  zf,  and  rf  =  xj  +  y?  +  z22, 
this  gives 


cos«1,  yi  =  1\  cos  fa,  Zi  = 
=  r2  cos  «2,  2/2  =  r2  cos  Pz>  zi  —  rz  cos  72- 


Now 
and 

Substituting, 

cos  0  —  cos  «!  cos  0-2  +  cos  Pi  cos  fa  +  cos  yx  cos  y2, 

a  relation  which  is  independent,  as  it  must  be,  of  the  particular 
points  Pl  and  P2  chosen. 

The  corresponding  formula  in  the  plane  for  the  angle  be- 
tween two  lines  is 


cos  0  =  cos  «!  cos  a%  +  sin  o^  sin 


462  UNIFIED  MATHEMATICS 

The  adaptation  and  the  proof  are  left  to  the  student  as  an 
exercise.  Compare  with  formula  on  page  247.  ' 

Frequently  the  direction  cosines  of  a  line  in  space  are  repre- 
sented by  I,  in,  n  or  11}  m^  nl7  etc.  The  condition  that  two 
lines  be  perpendicular  is  evidently  ?i/2  +  mlm<,  +  ?*1w2  =  0,  and 
the  condition  for  parallel  lines  is  that  I,  =  12,  ml  =  ra2,  and 
nt  =  n2.  It  can  be  shown  by  using  vectors  that  the  relation 
.  /!/2  4-  wiiWo  +  «i»2  =  1>  combined  with  If  +  mf  +  n^  =  1  and 
12-  +  w22  +  n22  =  1  reduces  to  lt  =  /,,  rat  =  ra2  ,  and  %  =  n». 

9.    First-  degree  equation.  —  We  now  show  that  the  equation 

(a)  Ax  +  By+Cz  +  D  =  Q 

represents  a  plane.  A  plane  is,  by  definition,  a  surface  which 
is  such  that  the  straight  line  joining  any  two  points  in  the 
surface  lies  wholly  on  the  surface,  i.e.  any  other  point  on  the 
line  is  also  on  the  surface. 

Let  PI(XI}  y},  Zx)  and  P2(x2,  y^,  z2)  be  any  two  points  which 
satisfy  equation  (a). 

(6)  .:Axl  +  Byi+Czl+D  =  0. 

(c)  ^la:2  +  By*  +  Cz»  +  D  =  0. 

Let  P3(%3,  2/3,  z3)  be  any  other  point  on  the  line  joining 
P(XI,  yi,  %i)  to  P2(a-2,  y2,  z2)  and  let  this  point  divide  P:P2  into 

P  P      Jc 

segments  such  that  ~—  -  =  —  • 

PS  PZ      ""2 
Then,  #3,  y3,  and  z3  may  be  written 

_ 
3  ~~ 


Substituting  these  values  in  the  left-hand  member  of  equa- 
tion (a),  we  have 


[  B       i          z  | 


fti  -f-  fc2  KI  -\-  K%  Kl  -|-  ft"2 

which  may  be  written,  by  rearrangement  of  terms, 

-A-  (Ax,  +  By,  +  &!+  D)  +  r^y-  (^2  +  By,  +  Cz2+D)  ; 


SOLID  GEOMETRY:  POINTS  AND  LINES       463 

but  this  expression,  by  (&)  and  (c)  above,  is  zero  ;  hence  the 
point  P3(x3,  1/3,  z3),  any  point  on  the  straight  line  joining  PI  to 
P2>  which  points  are  on  the  locus  of  (a),  is  also  on  this  locus. 

.-.  Ax  +  By  +  Cz  +  D  =  0  represents  a  plane. 

The  converse  proposition  is  demonstrated  in  section  5  of  the 
next  chapter. 

A  plane  to  pass  through  three  given  points  is  determined  by 
substitution  of  the  three  points  in  equation  (a)  and  solving  for 
three  of  the  constants,  e.g.  A,  B,  and  C,  in  terms  of  the 
fourth.  If  the  fourth  constant  chosen  happens  to  be  zero, 
another  selection  must  be  made. 

Illustrative  problem.  —  Find  the  plane  through  (4,  4,  4), 
(3,  0,  -  5),  and  (0,  -  3,  -  7). 

Let  Ax  +  By  +  Cz  +  D  =  0  represents  the  equation  of  the  plane. 
Substituting, 

x  3  (1)  4^l  +  4B  +  4C  +  .D  =  0. 

(2)  3  A  -  5  C  +  D  =  0. 

x  4  (8)         -3#-7C+D  =  0. 

Since  it  happens  that  only  A,  C,  and  D  occur  in  (2),  eliminate  B 
between  (1)  and  (3)  by  multiplying  (1)  by  3  and  (3)  by  4  and  adding, 
obtaining 

(4)  12  A  -  16  C  +  1  D  =  0. 

-4(2)     3A-    5  C+     J>  =  0. 

Eliminate  A  between  (4)  and  (2)  by  multiplying  (3)  by  —  4  and  add- 
ing to  (4).  This  gives 

(5)  4  C  +  3  D  =  0  ;   C  =  -    fD. 

Substituting  in  (2)  the  value  of  C  found  gives 

3  A  +  *£  D  +  D  =  0,  A  =  —  |f  D. 
Substituting  value  of  C  in  (3)  gives 


The  equation  of  the  plane  may  be  written, 

-  }f  Dx+  }  {  Dy  -  J  Dz  +  D  =  0,  or 

—  19x  +  26y  —  9z  +  12  =  0.    Ans. 


464  UNIFIED  MATHEMATICS 

PROBLEMS 

1.  Find  the  angle  between  the  two  lines  of  problem  5  of 
the  preceding  list  of  problems. 

2.  Find  the  direction  cosines  proportional  to  3,  —2,  and  —5  ; 
find  those  proportional  to  2,  3,  and  —4  ;  find  the  angle  between 
two  lines  having  these  direction  cosines  that  you  have  found. 

3.  Find   the   equation  of  a  plane  whose  intercepts  on  the 
axes  are,  respectively,  3,  —  5,  and  7. 

4.  Find  the  equation  of  a  plane  through  the  points  (3,  0,  5), 
(-2,  11,  7),  and  (0,  11,  7). 

5.  The  parametric  equations  of  any  line  through  (3,  0,  5) 
can  be  written 

x  —  3  =  r  cos  a,  y  —  0  =  r  cos  /3,  z  —  5  =  r  cos  y. 
If  this  line  is  to  pass  through  (—2,  11,  7),  these  coordinates 
must  satisfy  these  equations.  Make  the  substitutions,  re- 
spectively; square  and  add  the  corresponding  numbers  and 
thus  obtain  the  value  of  r,  giving  the  distance  of  the  point 
(3,  0,  5)  from  (—2,  11,  7).  Find  then  the  values  of  cos  a, 
cos  (3,  and  cos  y. 

6.  If  a  rectangular  box  with  sides  parallel  to  the  coordinate 
planes  has  the  line  joining  (3,  0,  5)  to  (—  2,  11,  7)  as  a  principal 
diagonal,  find  the  lengths  of  the   sides,   the   length   of   the 
diagonal,  and  so  find  the  direction  cosines  of  the  line  joining 
the  two  points. 

7.  Discuss  the  loci  of  the  following  equations  and  find  three 
points  on  each  locus  : 

a.  3  x  +  11  =  0. 

b.  z2-9  =  0. 

c.  z-5  =  0. 

d.  x  —  y  —  5  =  0. 

e.  z 


8.   Where  do  points  lie  which  are  common  to  the  loci  of  the 
two  following  equations  :         x  =  3,  and  y  =  5  ? 

3-3  =  0  and  2-2y  +  10  =  0? 


CHAPTER   XXXI 

SOLID  ANALYTICS;  FIRST-DEGREE  EQUATIONS  AND 
EQUATIONS   IN   TWO   VARIABLES 

1.  Locus   of  an   equation  in  three  variables.  —  Any  equation 
involving  three   variables  has  for  its   locus  a   surface  which 
may,  in  special  forms  of  the  equation,  reduce  to  one  or  more 
lines  or  points.     We  obtain  points  on  such  a  surface  by  giving 
values  to  two  coordinates,  e.g.,  x  and  y,  and  solving  for  the 
third,  e.g.,  z.     Thus  we  have  found  that  any  first-degree  equa- 
tion represents  a  plane. 

NOTE,  x2  +  y2  +  «2  =  0  represents  only  a  point  (0,  0,  0),  or  a  point 
sphere. 

x2  +  y2  =  0  represents  the  z-axis  since  everywhere  on  this  axis  x  =  0 
and  y  =  0. 

2.  Intersections  of  loci.  —  (See  Chapter  V,  Section  2.)     Any 
point  which  satisfies  two  equations  involving  three  variables 
lies,  in  general,  upon  a  curve  which   is  common  to  the  two 
surfaces  represented. 

When  three  equations  are  regarded  as  simultaneous,  points 
of  intersection  of  the  three  surfaces  are  obtained.  Under 
special  relations  between  the  three  given  equations,  these 
points  may  lie  upon  a  line,  but,  in  general,  three  simultaneous 
rational  integral  algebraic  equations  determine  a  finite  num- 
ber of  points  of  intersection. 

Just  as  a  family  of  lines  through  the  intersection  of  two 
given  lines  is  obtained  in  Chapter  V,  Section  4,  in  the  form 
lj_  -f  kL  =  0,  so  the  equation  f^x,  y,  z)  +  kf2(x,  y,  z)  =  0 

465 


466 


UNIFIED  MATHEMATICS 


represents  a  family  of  surfaces  which  pass  through  the  inter- 
section curves  of  the  two  given  surfaces. 

Thus,  z2  +  j/2  +  z2  =  25  represents  a  sphere  of  radius  5  ;  z2  =  9  repre- 
sents two  planes  parallel  to  the  yz-plane  ;  the  equation 
X2  +  yz  +  Z2  _  25  -  fc(z2  -  9)  =  0 

represents  for  all  values  of  A;  a  surface  through  the  intersections  of  the 
sphere  and  the  plane.     For  k  =  —  1,  this  surface  reduces  to  a  cylinder, 


3,2 


Z2  _  16  =  0. 


3.  Cylindrical  surfaces.  —  Any  equation  in  two  variables,  as 
x  and  y,  represents  in  space  a  cylinder  whose  axis  is  parallel 
to  the  axis  designated  by  the  third  variable. 


Cylindrical  surfaces: 

ELLIPTIC  HYPERBOLIC  PARABOLIC 

The  curves  indicated  on  these  surfaces  are  cubic  space  curves. 

If  an  equation  f(x,  y)  =  0  is  given  in  x  and  y,  any  point 
(xi>  2/i)  which  satisfies  the  equation  will  lie  upon  the  curve  in 
the  xy-pla,ne  given  by  f(x,  y)=  0.  Considered  as  a  point  in 
space,  the  point  (x1}  y1}  0)  satisfies  the  equation,  and  further 
it  is  evident  that  (ajl5  y1}  z),  irrespective  of  the  value  of  z,  will 
also  satisfy  f(x,  y)  =  0,  since  the  z-coordinate  does  not  enter 
at  all. 


SOLID  ANALYTICS 


467 


Thus,  (3,  4,  0)  satisfies  the  equation 

x2  +  y2  —  25  =  0 

and  also  the  points  (3,  4,  1)  or  (3,  4,  —  10)  or  (3,  4,  8,)  will  satisfy  the 
equation 

/^2    l    i*2 y^  (i 

But  all  points  (xi,  y1}  z),  for  varying  values  of  z  only,  lie  on  a 
parallel  to  the  z-axis  through  (xt,  y^  and  hence  all  points  on 
the  surface  generated  by  a  straight  line  moving  parallel  to  the 
z-axis  and  touching  the  curve  f(x,  y)  =  0,  in  the  ary-plane,  lie 
upon  a  cylinder.  The  curve  f(x,  y}  =  0  is  called  the  directrix 
of  the  cylinder  and  the  moving  line  is  called  the  generator  or 
element  of  the  cylindrical  surface.  Similarly,  when  an  equa- 
tion is  given  in  x  and  z  or  in  y  and  z,  a  cylinder  is  represented. 
A  plane  given  by  a  first-degree  equation  in  two  variables, 
or  one  variable,  is  a  special  case  of  the  preceding. 

4.    Straight  line  as  the  intersection  of  two  planes.  —  Just  as  the 
equation  ^ — —  =  —    — -  represents  in  the  plane  the  straight 

line  joining  PJ(XI,  .Vi)  to  P2(xz,  yz)>  so  the  three  equations, 
x  —  Xi y  —  yi z  —  Zi 


represent  in  space  the 
straight  line   joining 


There  are  three  equal- 
ities which  are  ob- 
tained by  leaving  out 
in  turn  each  of  the 
fractions,  but  there 
are  only  two  inde- 
pendent equations,  as 


v 


Line  joining  PI  to  P-  in  space 


468  UNIFIED  MATHEMATICS 

the  third  equality  would  follow  always  from  the  first  two 
which  were  given. 

These  formulas  can  be  obtained  directly  from  the  properties 
of  similar  triangles,  or  from  the  parametric  equations  of 
section  5  of  the  preceding  chapter.  The  latter  method  brings 
out  the  important  fact  that  the  values  x2  —  x^  yz  —  y^  and 
z2  —  Zi  are  proportional  to  the  direction  cosines  of  the  given 
line,  and  the  values  themselves  of  these  cosines  can  be 
obtained,  using  the  fact  that  the  sum  of  the  three  squares  is 
equal  to  unity.  The  derivation  of  the  theorems  mentioned  is 
left  as  an  exercise  to  the  student. 

The  parametric  forms  of  the  equations  of  a  straight  line  may 
be  written, 

fl-fli  _  y_-J/i  _  z-zl  _  r 
cos  a       cos  /8      cos  y 
or  \ 

a;  — a?!  _  y  —  y\  _  z  —  z^  _r 
k  cos  a     k  cos  (3     k  cos  y     k 

Further,  any  equations  which  can  be  put  in  one  of  the  two 
forms  above  represent  a  straight  line,  and  the  denominators  of 
the  fractions  are  proportional  to  the  direction  cosines  of  the 
line. 

The  equations  of  the  straight  line  in  the  form 

x  — -  Xi  _  y  —  yi  _  z  —  Zi 
a  b  c 

wherein  a,  6,  and  c  are  necessarily  proportional  to  the  direction 
cosines  of  the  line,  are  called  the  standard  or  symmetrical 
equations  of  the  line. 

In  general,  any  curve  in  space  is  given  as  the  intersection  of 
two  surfaces  by  the  equations  of  the  two  surfaces.  In  particu- 
lar, the  straight  line  is  given  by  the  equations  of  any  two 
planes  which  pass  through  the  line.  Of  the  infinite  number 
of  planes,  the  pencil  of  planes,  which  pass  through  a  given 
line,  the  three  planes,  called  projecting  planes  of  the  line, 
which  are  parallel  to  the  coordinate  axes  are  of  particular 


SOLID  ANALYTICS  469 

importance.  These  equations  will  evidently  be  first-degree 
equations  in  two  variables.  In  the  standard  form  the  equality 
of  any  two  members  gives  one  of  the  projecting  planes  through 
the  given  line. 

Illustrative  problem.  —  Find  the  direction  cosines  of  the 
straight  line  determined  by  the  two  planes 

(a)  x  +  y-3z-   5  =  0. 

(6)  3  x  -  y  -  5  z  -  11  =  0. 

Find  the  projecting  planes  parallel  to  the  coordinate  axes  (or .perpen- 
dicular to  the  coordinate  planes)  ;  find  the  points  where  this  line  pierces 
the  coordinate  planes. 

Any  plane  through  the  line  of  intersection  is  given  by 

(c)  x  +  y-3z-5  +  fc(3x-y-5z-ll)=0. 

Giving  to  k  the  value  —  f,  which  is  equivalent  to  multiplying  (a)  by  5, 
and  (6)  by  —  3  and  adding,  and  simplifying  you  have, 

5  x  +  5  y  -  15  z  -  25  -  9  x  +  3  y  +  15  z  +  33  =  0,  or 

(d)  —  4x  +  8j/  +  8=0,  as  the  plane  of  projection  on  the  xy-plane. 
Eliminating  y,  k  —  1,  gives, 

(e)  4  x  —  8  z  —  1C  =  0  or  x  —  2z  —  4  =  0,  the  plane  of  projection  on 

the  xz-plane. 

Eliminating  x,  k  =  —  £,  gives, 

(/)   _4(/  +  4z-f4  =  0,   which  might  have  been  obtained  from  (d) 
and  (e),  thepiane  of  projection  on  the  yz-plane. 
Solving  for  x,  in  (d)  and  (e), 

x  =  2(y  +  1)  and 
x  =  2(z  +  2). 
x  =  2(j/+  l)=2(z  +  2). 
x-0_y+ l_z  +  2 

2  1  1 

The  denominators  2,  1,  and  1  are  proportional  to  the  direction  cosines  of 
this  line.  Hence 

cos  «  =  2  m,  cos  /3  =  1  m  and  cos  7  =  1  m, 
giving, 

cos2  a  +  cos2  /3  +  cos2  -y  =  4  m2+  mz-f-  m2=  1;  6m2=l;  m=;±  — -  . 

V6 
Either  sign  may  be  taken,  but  for  convenience,  make  cos  «  positive. 

oil 

cos  a  =  — ^ ,  cos  p  =  — ^ ,  cos  y  =  — ^ ,  the  direction  cosines. 
V6  V6  V6 


470 


UNIFIED  MATHEMATICS 


To  find  where  this  line  pierces  any  coordinate  plane,  as  z  =  0,  solve 
the  equation  of  the  coordinate  plane  as  simultaneous  with  the  two  given 
planes  which  determine  the  line. 

This  gives  here  (4,  1,0)  as  the  piercing  point  with  the  xy-plane. 
Similarly  we  find  the  intersection  with  any  plane. 

A  parallel  line  to  our  given  line  through  a  given  point  would  be  de- 
termined by  two  planes  through  the  given  point  parallel  to  the  two  given 
planes  which  determine  the  line.  Why  ?  Determine  the  parallel  line 
through  (1,  —  5,  6). 

5.  Normal  form  of  the  equation  of  a  plane.  —  (See  Section  3, 
Chapter  IX.)  In  the  plane,  the  equation  x  cos  a+y  sin  a—p=Q, 
which  may  be  written  x  cos  a  +  y  cosft—p  =  0,  represents 
the  equation  of  a  straight  line  in  normal  form,  which  line  is 
such  that  the  perpendicular  from  the  origin  upon  it  has  the 
length  p  and  makes  the  angles  a  and  ft  with  #-axis  and  y-axis. 
Similarly,  in  space,  the  equation 

x  cos  «  +  y  cos  ft  +  z  cos  y  —  p  =  0 

represents  a  plane  which  is  such  that  the  perpendicular  from 
the  origin  upon  it  has  the  length  p  and  makes  the  angles  «, 

ft,  and  y  with  the 
ic-axis,  ?/-axis,  and 
2-axis  respectively. 

Evidently,  if  a 
plane  is  given  and 
a  perpendicular  ON 
of  length  p,  having 
direction  cosines  a, 
ft,  and  y,  is  dropt 
from  the  origin  to 
this  plane,  the  point 
Nis  (p  cos  a,  p  cos  ft, 
p  cos  y)  and  the 
extension  of  the  per- 


ON  of  length  p;  ON'  of  length  2p 
Direction  angles  of  ONN' :  a,  P,  -y 


pendicular  by  the  length  p  gives  the  point  N'  (2  p  cos  a, 
2p  cos  ft,  2p  cos  y).  Any  point  P(x,  y,  z)  which  is  equidis- 
tant from  0(0,  0,  0)  and  N'(2pcosa,  2  p  cos  ft,  2  p  cosy) 


SOLID  ANALYTICS  471 

lies  on  our  plane.     Writing  and  equating  these  distances,  we 
have, 

a2  +  y1  +  z2  =  (x  -  2  p  cos  «)2  +  (y  -  2  p  cos  /3)2  +  (z  -  2  p  cos  /3)2. 
Whence, 


cos  «)#  +  (4p  cos  /3)i/  +  (4p  cos  y)z 

=  4  J92(COS2  «  +  COS2  ft  +  COS2  y), 

giving  finally 

x  cos  a  +  y  cos  /8  -f-  z  cos  y  —  p  =  0 
as  the  equation. 

In  the  plane  the  distance  from  any  point  (a^,  y^  to  a  line  is 
obtained  by  writing  the  equation  of  the  line  in  normal  form  and 
substituting  therein  for  x  and  y,  xt  and  yv  In  space  the  dis- 
tance of  a  point  (x1}  y1}  Zj)  from  a  plane  is  obtained  by  writing  the 
equation  of  the  plane  in  normal  form  and  substituting  therein 
these  coordinates  of  the  point  for  x,  y,  and  z,  respectively. 

To  reduce  a  linear  equation  to  normal  form,  you  divide  the 
equation  through,  after  transposing  all  terms  to  the  left-hand 
member,  by  the  square  root  of  the  sum  of  the  squares  of  the 
coefficients  of  x,  y,  and  z,  choosing  the  sign  opposite  to  the  sign 
of  the  constant  term.  The  proof  is  not  similar  to  the  proof 
of  the  corresponding  theorem  in  plane  analytics. 

Parallel  planes  are  represented  by  linear  equations  having 
the  corresponding  coefficients,  of  x,  y,  and  z,  equal  or  propor- 
tional. 

PROBLEMS 

1.  Put  the  following  equations  in  normal  form  and  deter- 
mine the  distance  of  each  plane  from  the  origin  : 

a.   2x-3y+4:Z-ll=0. 

6.   a;-f-y  +  z-5  =  0. 

c.   2  x  -  3  y  —  11  =  0.        d.  z-7  =  Q. 

Determine  the  direction  cosines  and  the  direction  angles  of 
the  normals  to  each  of  the  above  planes. 


472    -  UNIFIED  MATHEMATICS 

2.  Find   the   equations   in   standard  form   of  a  line  from 
(1,   2,   5)    perpendicular    to   the   first    plane   in   problem   1 ; 
through  (0,  0,  0)  perpendicular  to  the  second  plane  in  problem 
1;  through  (—2,  —  3,  4)  perpendicular  to  the  third  plane  in 
problem  1.     Determine  in  each  of  these  three  problems  the 
intersection  of  the  perpendicular  with  the  plane. 

3.  Find  the  piercing  points  with  the  coordinate  planes  of 
the  following  lines : 

a.  2  x  —  3  y  +  4  2  —  11  =  0  and  x  —  y  +  z  —  5  =  0. 

b.  2x—  3y+4z-ll  =  0andz  —  7  =  0. 

c.  2z-3y- 11  =  0  and  z-7  =  0. 

4.  Put   the   three   lines   of   problem   3  in   standard   form. 
Note  that  in  the  second  and  third  cases,  since  the  given  line 
lies  in  a  plane  parallel  to  the  x?/-plane,  the  line  makes  an  angle 
of  90°  with  the  z-axis,  i.e.  cos  y  =  0.     The  equations  of  the  sec- 
ond and  third  lines  in  standard  form  would  have  a  zero  de- 
nominator, and  so  it  is  better  to  put  these  equations  in  the  form 
given  in  the  third  of  these  problems.     The  values  of  cos  a  and 
cos  ft  are  determined  here  from  the  equation  2  x  —  3y  —  11  =  0, 

2  3 

giving  cos  a  =  —  —  and  cos  ft  = 


V13  V13 

5.  Find   the   equations  of   the   straight  lines   through  the 
two  points  :        a.    ^,  5,  -  2)  and  (0,  0,  7). 

6.    (3,  5,  -  2)  and  (0,  0,  0). 

c.    (3,  5,  -  2)  and  (-  3,  5,  +2). 

6.  Find  the  angle  between  the  lines 

d*-2_y_z-l 
'  ~~ 


7.  Do  the  two  lines  in  problem  6  intersect  ?  How  can  you 
determine  whether  any  two  given  lines  intersect  ?  Note  that 
the  problem  is  entirely  analogous  to  the  problem  in  plane 
analytics  as  to  whether  three  given  lines  intersect,  and  is 
solved  in  the  same  manner.  Write  the  equations  of  two 
intersecting  lines. 


SOLID  ANALYTICS 


473 


8.   Determine  the  curve  in  which  the  sphere 

x*  +  f-  +  z2  —  400  =  0 

is  intersected  by  the  plane  y  —  9  =  0.     Note  that  substituting 
y  =  9  is  equivalent  to  writing 


Sphere  cut  by  a  plane 

which  gives,  of  course,  a  new  surface  passing  through  the 
intersection  curve  of  the  first  two  surfaces. 

9.   Find  the  intersection  of  the  sphere  x2  +  y2  +  zz  — 100  =  0 
and  the  cylinder  x2  +  y2  —  36  =  0. 

10.  Upon  what  cylinder,  parallel  to  one  of  the  coordinate 
axes,  does  the  intersection  of  the  plane  x  =  5  with  the  surface 
ar2  +  4  y2  =  25  z  lie  ? 

11.  Find  the  intersection  of  the  line 

with  the  sphere  x2  +  y-  +  z2  — 100  =  0  by  substituting  these 
values  in  the  equation  of  the  sphere  and  solving  for  r.  Note 
that  since  the  right-hand  coefficients,  2,  3,  and  —  5,  are  not  the 


474  UNIFIED  MATHEMATICS 

direction  cosines  of  this  line,  but  only  proportional  to  them, 
the  values  of  r  obtained  are  not  the  distances  from  (1,  2,  —  3) 
to  the  points  of  intersection  with  the  sphere,  but  are  propor- 
tional to  these  distances.  The  points  of  intersection  are 
obtained  by  substituting  the  values  of  r  found  back  in  the 
equations  of  the  line  and  solving  for  (x,  y,  z). 


Hyperboloid  of  one  sheet  Hyperboioid  of  two  sheets 

CHAPTER   XXXII 

SOLID  ANALYTICS:    QUADRIC   SURFACES 

1.  General  equation.  —  In  plane  coordinates,  any  equation  of 
the  second  degree  represents  a  conic  section.  Similarly,  in 
space  coordinates,  any  equation  of  the  second  degree  repre- 
sents a  quadric  surface.  The  types  of  quadric  surfaces, 
limited  in  number,  are  closely  allied  to  the  types  of  conic  sec- 
tions. In  plane  analytics,  it  is  shown  that  the  general  equation 
of  the  second  degree,  containing  the  "  cross-term "  xy,  intro- 
duces no  new  curves,  only  the  same  curves,  represented  by 
the  different  types  of  equations  in  which  no  xy-term  appears. 
It  is  likewise  true  in  space  that  the  general  equation  contain- 
ing any  or  all  of  the  "  cross-terms,"  yz,  xz,  and  xy,  presents 
no  surfaces  different  from  those  which  may  be  represented 
by  the  general  equation  containing  no  cross-term. 

Methods  of  transformation  of  coordinates  quite  similar  to 
those  discussed  in  Chapter  XXIV  apply  to  space  coordinates, 
but  the  limitations  of  a  first  course,  preclude  any  discussion  of 
the  methods  and  results. 

Any  surface  given  by  an  equation  of  the  second  degree  is 

475 


476 


UNIFIED  MATHEMATICS 


cut  by  any  plane  in  some  form  (including,  of  course,  limiting 
forms)  of  conic  section.  The  coordinate  planes  very  evidently 
cut  any  quadric  surface  in  a  conic,  since  the  curve  of  inter- 
section in  the  coordinate  plane  is  given  by  an  equation  of  the 
second  degree  in  the  two  variables  of  that  plane.  The  trans- 
formations mentioned  above  are  desirable  for  the  general  proof, 
but  another  method  is  indicated  below. 


SPHERE 


Ellipsoids : 

PROLATE  OBLATE     .         GENERAL 

SPHEROID  SPHEROID  ELLIPSOID 


2.   Ellipsoids.  —  The  equation  of  a  sphere  has  been  given  as 


An  ellipsoid  is  given  by  the  equation 


a2  62 

This  surface  is  related  to  the  three  spheres, 


(X  - 


_  7)2  = 


very  much  as  the  ellipse  is  related  to  its  auxiliary  circles. 
The  parametric  equations  of  the  above  ellipsoid  are 

x  —  h  =  a  cos  a, 
y  —  Jc  =  b  cos  (3, 
z  —  I  =  c  cos  y. 


SOLID  ANALYTICS:  QUADRIC  SURFACES       477 


The  elimination  of  a,  (3,  and  y,  employing 

cos2  a  +  cos2  ft  +  cos2  y  =  1, 

gives  the  equation  of  the  ellipsoid  in  the  standard  form  above. 

The  quantities  a,  b,  and  c  represent  the  semi-axes  of  the 

ellipsoid.     If   two  of  these  denominators  are  equal  to  each 


Ellipsoid  of  revolution,  with  x-axis  as  axis  of  revolution 

other,  the  ellipsoid  is  an  ellipsoid  of  revolution  about  an  axis 
parallel  to  the  axis  corresponding  to  the  term  with  the  odd 
denominator. 
Thus, 


25     16     16 


is  an  ellipsoid  of  revolution,  obtained  by  revolving  the  curve  --  (-      =  1 
about  the  x-axis. 

The  derivation  of  the  formula  of  the  ellipsoid  of  revolution 
is  as  follows,  PN*  +  NM2=PM*,  but  QM=PM,  radii  of 
the  circle  QPR  about  M,  with  lettering  as  indicated  on  diagram 
given  above.  Now  for  all  points  on  this  circle  the  ^-coordinate 
is  the  same, 


25 

which  is  a  relation  true  for  every  point  on  the  circle  obtained 


478  UNIFIED  MATHEMATICS 

by  rotating  the  point  Q  on  the  ellipse  -™  +  ^—  =  1  about  its 

25      16 

axis.  But  Q  is  a,ny  point  on  the  ellipse,  and  hence  P  may  be 
any  point  on  the  surface  obtained  by  revolving  the  ellipse 
about  its  axis. 

Hence,  for  every  point  on  this  surface, 

,*  +  **  =  16(1- 
or 

*!  4-^.4.11  =  1 

25     16     16 

Any  ellipsoid  of  revolution  obtained  by  revolving  an  ellipse 
about  its  major  axis  is  called  a  prolate  spheroid,  and  is  shaped 
like  a  football ;  an  oblate  spheroid  is  obtained  by  rotating  an 
ellipse  about  its  minor  axis,  and  is  shaped  like  a  circular 
cushion  or  the  earth. 

PROBLEMS 

1.  Find  the  equation  of  the  Iphere  having  the  center  at  the 
origin  and  passing  through  the  point  (—2,  5,  6).     Give  the 
seven  points  which  lie  on  this  sphere  and  are  symmetrically 
situated   to  the   given  point  with  respect   to   the   coordinate 
planes. 

2.  Find  the,  equation  of  the  preceding  sphere  if  the  center 
is  at  (3,  —  2, 12).     Find  by  using  conditions  of  symmetry  with 
respect  to  planes  through  the  center  parallel  to  the  coordinate 
planes  seven  further  points  on  this  sphere. 

3.  Write  the  equation  of  the  ellipsoid  having  the  center  at 
the  origin  and  semi-axes  equal  to  2,  3,  and  5  respectively  (x,  y, 
and  z  order).     Find  three  points  on  this  ellipsoid.     Write  the 
equations  of  three  circles  which  lie  on  this  surface.     Write  the 
equations  of  the  traces  on  the  coordinate  planes,  i.e.  the  inter- 
sections with  these  planes.     Draw  the  graph. 

4.  If  a  football  is  10  inches   long  with  a   diameter  of   8 
inches,  write  the  equation  of  the  surface,  assuming  it  to  be  an 
ellipsoid.     Draw  the  graph  to  scale. 


SOLID  ANALYTICS:  QUADRIC  SURFACES       479 

5.  Assuming  that  an  air  cushion  18  inches  in  diameter  and 
6  inches  high  is  an  ellipsoid,  write  the  equation  of  the  surface. 
Draw  the  graph  to  scale. 

6.  Find  the  six  principal  foci  of  the  ellipsoid  in  problem  3. 
These  are  the  foci  of  the  traces  on  the  coordinate  planes. 

3.   Hyperboloids.  —  By  rotating  the  hyperbola 


about  either  axis,  a  hyperboloid  of  revolution  is  obtained. 
Rotation  about  the  principal  axis,  the  #-axis  here,  gives  a 
surface  of  two  separated  parts,  called  a  hyperboloid  of  revolution 
of  two  sheets.  The  equation  is, 


_ 
a2     62     62  ~ 

The  method  of  derivation,  which  we  outline,  is  general,  and 
being  applied  to  the  surface  obtained  by  revolving  any  curve, 
y  =/(aO  about  the  cc-axis,  will  give  the  equation  of  the  surface 
in  the  form  y2  +  z2  =  [/(z)]2. 

.  aj2  yZ 

Given  --  7-  =  1>  revolved  about  the  «-axis. 
a2     o2 


Hyperboloid  of  two  sheets,  of  revolution 
The  curves  are  slightly  distorted. 


480 


UNIFIED  MATHEMATICS 


Any  point  P(x,  y,  z)  on  this  surface  is  obtained  by  the  rotation 
of  a  point  Q(x,  y,  0)  about  the  x-axis. 

The  point  Q  generates  a  circle  in  a  plane  parallel  to  the  yz- 
plane,  in  which  x  has  everywhere  the  value  given  by  OM. 

The    equation   of    this 
circle  is 


Hyperboloid  of  two  sheets 
y-axis  as  principal  axis. 


This  radius  r  is  evidently 
a  function  of  x,  being 
defined  by  the  original 
equation  given ;  hence, 

rz  =  (ordinate  on  the  hy- 
perbola)2 =  bz(—  - 1 V 

V«2     / 

Hence  the  point  P(x,  y,  z) 
satisfies  the  equation 


or,  by  rearrangement   of 
terms, 


the  hyperboloid  of  revolu- 
tion of  tico  sheets. 

Note  that  precisely  this 
surface  would  have  been 
obtained  by  revolving 


a;2 


- —  —  s=  1  in  the  orz-pla.ne  about  the  x-axis. 
a2     62 

Similarly,  the  hyperboloid  of  revolution  of  one  sheet  is  obtained 
by  revolving  a  hyperbola  about  its  conjugate  axis.  The  pre- 
ceding hyperbola  revolved  about  the  y-axis  gives 


SOLID  ANALYTICS:  QUADRIC  SURFACES       481 


__ 
a2     &2     a2 

The  student  will  note  that  the  axis  of  rotation  in  each  case 
is  given  by  the  odd  term. 

Corresponding  to  these 
surfaces  of  revolution  are 
the  general  hyperboloids, 


a2      62 

hyperboloid  of  two  sheets, 
and 

^_^2  +  ?_2  =  l 
a2      62     c2 

hyperboloid  of  one  sheet, 

which  represent  in  each 

case  a  surface  having  a 

principal  axis  parallel  to 

the  axis  of  the  odd  term, 

e.g.  the  first  has  the  cc-axis 

as  principal  axis  and  the 

second  has  the  y-axis  as 

principal    axis.      Chang- 

ing the  principal  axis  to  another  of  the  coordinate  axes  inter- 

changes two  of  the  algebraic  signs  in  the  equation. 

4.  Paraboloids.  —  By  revolving  the  parabola  yz  =  4  ax  about 
its  axis  the  surface  \f  +  z"-  =  4  ax  is  obtained. 

This  is  called  a  paraboloid  of  revolution,  or  a  circular  parabo- 
loid, and  is  the  type  of  surface  which  is  fundamental  in 
theater  and  auditorium  construction.  The  derivation  of  the 
equation  is  left  as  an  exercise  for  the  student. 

The  elliptic  paraboloid  is  given  by  the  equation 


Hyperboloid  of  revolution,  one  sheet 
y-axis  as  principal  axis. 


and  sections  parallel  to  the  icy-plane  are  ellipses. 


482 


UNIFIED  MATHEMATICS 


Elliptic  paraboloid 


Hyperbolic  paraboloid 


;      -—-—5^ 

The    corresponding 

»"  IL  "             standard     forms      with 

J  _ 

-  jj  II 

-5-"--"                                         A            '4-V, 

/ 

::::! 

I  1                                                 ,-        1                             -        - 

rrk 

*<:: 

--]  --            respectively,    as    prmci- 

V 

s:' 

A~             i      * 

± 

/ 

/                  ^ 

----  + 

V^ 

~T* 

—• 

S 

^=kz 

1 

—  ;                            r2     z2 

^r4- 

::/::                         -,  +  -^  =  4.V> 

\ 

-    -    -  -                                  tt2        C2 

/ 

\ 

/ 

\ 

c:                             oo 

:: 

... 

:±                                  0s      C* 

:    ::  _ 

t- 

i 

J_IT 

/ 

-  [- 

/ 

3 

7 

t 

;                t 

y   " 

—  —  ^Z  —  —  —  Q  % 

j 

:::jL:: 

a2      62 

S                    '       r^"               ' 

'M# 

•JTT^JBJI:       gives  the   most   compli- 
cated  of  the   quadratic 

E 

nip 

tic  p 

ar 

al 

ill 

f 

£::::::::::i 
joloid  of  revc 

surfaces,   the   hyperbolic 
paraboloid,     a      saddle- 
lution              shaped    surface     which 

SOLID  ANALYTICS:  QUADRIC  SURFACES       483 

is  here  represented  by  a  photograph  of  a   model  of  such  a 
surface. 

The  hyperbolic  paraboloid  may  be  generated  by  the  motion 
of  a  given  parabola,  x2  —  4  a2z  =  0,  moving  parallel  to  the  xz- 
plane  and  having  its  vertex  moving  on  the  parabola 


5.  Cones.  —  If  any  straight  line  is  revolved  about  another 
straight  line  in  its  plane  as  an  axis,  a  cone  of  revolution  is 
generated.  Limit- 
ing forms  are  the 
cylinder,  when  the 
re vol  viiig  line  is  par- 
allel to  the  axis,  a 
plane  when  the  re- 
volving line  is  per- 
pendicular to  the 
axis,  and  a  straight 
line  when  the  revolv- 
ing line  coincides 


Cone  generated  by  a  straight  line  rotating 

about  an  axis  in  the  same  plane 
with  its  axis. 

Let  v  =  -x  revolve  about  OX.     The  cone  of  revolution  geu- 


crated  has  the  equation 


The  cone  is  itself  a  limiting  form  of  the  hyperboloids,  as 
will  be  noted  below. 

The  general  equation  of  the  cone,  whose  axis  is  the  y-axis 
and  whose  vertex  is  the  origin,  is 


6.   Conic  sections.  —  The  method  which  we  will  here  outline 
to  prove  that  every  plane  section  of  a  cone  may  be  given  by 


484 


UNIFIED  MATHEMATICS 


Conic  sections : 


ELLIPSE 


I'AKA  HULA         HTFKKBOLJ 


an  equation  of  the  second  degree  applies  to  a  cone  having  an 
ellipse,  parabola,  or  hyperbola  as  a  base  as  well  as  to  the  cir- 
cular base,  which  is  taken  for  convenience. 


'*"         '/"        Z~ 

Given  the  cone  ^  =  0. 
a-      IP      b~ 

Evidently,  any  one  of  the  planes  given  by  x  =  k  cuts  this 

|||||i||ii^iiNiiiiiiiiiiiiiiiimtimimuu< 

~3J^ 
--it}<f- 

IliiiiliEE; 

-+3rjrx  — 

Si             //! 

::::i±::::: 

X^fefr/fc 

=41- 

—  - 

-:     i  srf  Tt:^~ 

-^3" 

jx^  1  /  ir^jrh 

::::::!::: 

^C-        /    j    ,  ! 

±  .  _. 

0  ^         /   /  /  \ 



^-?r 

-y  \.  .  .        v,  ./-  .\*-..~ 

.  r 

"4: 

II 

T-~nf  liii    1        iTfrHJA   — 
/i      '                          '  ^^^ 

M 

::;ff::ffi::::::::::::x|:;: 

~X/  — 

^•*-  •-                          — 

Cone  cut  in  an  ellipse  by  a  plane 
cone  in  a  circular  section,  with  a  point  as  limiting  case  when 


SOLID  ANALYTICS:  QUADRIC  SURFACES       485 


The  planes  y  =  k  and  z  =  k  cut  this  cone  in  hyperbolas. 
When  y  =  0  or  z  =  0,  the  hyperbola  "  degenerates  "  into  two 
straight  lines  intersecting  at  the  vertex. 

The  plane  y  =  mx  +  k  cuts  the  cone  in  a  curve,  which  we 
will  refer  in  this  plane  to  the  line  y  =  mx  +  A:,  intersection  of 
the  ccy-plane  and  the  cutting  plane,  and  the  line  y  =  k,  the 
intersection  of  the  t/z-plane  and  the  cutting  plane,  as  axes 
(a/  and  z')  of  coordinates.  From  any  point  P(x,  y,  z)  on  the 
curve  of  intersection,  drop  a  perpendicular  PM  to  the  xy- 
plane.  The  intersection  curve  satisfies  the  equation 

a?  _  (mx  +  fc)2  _  2?  _  0 
a2  fe2  W  ~ 

or  bn-x>  -  a\mx  +  fc)2  —  a2z2  =  0. 

Evidently,  PM  =  z  =  z',  since  PM  is  drawn  in  one  of  two  per- 
pendicular planes,  and  perpendicular  to  the  other. 


The  elliptic  section  depicted  in  its  own  plane 


Further,  BM=yf  = 

stituting,  we  have, 
6V2 
1  -f-m2 


-  a2m2)z'2  -  a2z  ' 


-|-  m2,  whence  x  = 
2  a*km 


x/ 


Sub- 


m2 


Vl  +  m2 
m2)- 


x  -  a=fc2  -  aV2  =  0, 


=  0. 


486 


UNIFIED  MATHEMATICS 


But  this  is  an  equation  of  the  second  degree  in  x'  and  z'.    Fur- 
ther, the  coefficient  of  x'2  is  (bz  —  a2m2)  and  of  z'2  is  —  a2(l-f  m2); 

hence  the  curve  is  an  ellipse  if  m2  >  — ,  a  parabola  if  m2  =  — , 

7i2  ^  ^ 

and  a  hyperbola  if  ra2  <  —  • 
a2 

When  the  cutting  plane  has  the  form  y  =  mx  +  nz  -f  k,  the 
proof  is  more  complicated  but  not  essentially  different. 

Every  section  of  a  cone  may  be  represented  by  an  equation  of 
the  second  degree  in  two  variables. 


Ellipse,  parabola,  hyperbola,  and  two  straight  lines  as  intersections  of  a 
cone  by  a  plane 

PROBLEMS 
1.   Name  and  discuss  the  following  surfaces : 

a.  x*+    y*+    z2  — 100  =0. 

b.  a2  +  2?/2  +  3z2-100  =0. 

c.  a2  +  2y2-4z2-100  =0. 

d.  x*  +  2y*  -100  =0. 

e.  x2  +  2y*-±z*  =0. 
/.    x2-2?/2-4z2-100  =0. 
g.   x2                      -100  =0. 
h.  x*-22 


SOLID  ANALYTICS:  QUADRIC  SURFACES       487 

i.  o;2-2.y2.  =0. 

j.  x-  +  2y2  =0. 

k.  z2  +  2  1/2  +  4  z2  =0. 

I  z2  +  2y2  +  4z2  +  100  =0. 

2.  How  would  the  addition  of  a  term,  10  x,  affect  the  locus 
of  each  of  the  preceding  twelve  expressions  ?     Discuss  the 
change  in  each  locus  produced  by  changing  the  sign  of  x2  in 
each  expression  from  +  to  —  . 

3.  Find  the  equation  of  the  cone  obtained  by  revolving  the 
line  in  the  a^/-plane  y  =  4  x  —  10  about  the  o>-axis  ;  about  the 
y-axis  ;  about  the  z-axis. 

4.  Find  the  equation  of  the  paraboloid  obtained  by  revolv- 
ing the  parabola  z2  =  8  x  about  the  #-axis  ;  find  the  equation  of 
the   surface    obtained   by   revolving   this    surface    about   the 
z-axis.     Why  has  the  latter   surface   not  received  particular 
discussion  ? 

5.  Find  the  equation  of  the  cone  obtained  by  revolving  the 
line  y  =  4  x  —  10  about  the  line  y  =  6.     Note  that  this  differs 
from  the  problems  which  we  have  considered  in  the  text  only 
by  a  change  of  origin,  or  a  transformation  of  the  type 


6.  Find  the  locus  of  a  point  which  is  equidistant  from  the 
point  (4,  0,  0)  and  from  the  plane   x  +  4  =  0.     What   is  the 
surface  ? 

7.  Find  the  locus  of  a  point  the  sum  of  whose  distances 
from  the  points  (4,  0,  0)  and  (  —  4,  0,  0  )  is  constant  and  equal 
to  10.     What  is  the  surface  ? 

8.  Find  the  locus  of  a  point  the  difference  of  whose  distances 
from  two  points  (4,  0,  0)  and  (—  4,  0,  0)  is  constant  and  equal 
to  6. 

9.  How  would  you  find  in  space  coordinates  the  distance 
from  a  point  to  a  line  ?     Apply  to  finding  the  distance  from 

PAJ.    i.u    v       x  —  2     y  —  3     z  —  8 
(1,  3,  —  5)  to  the  line  —  —  —  =  •-  —  —  =  —  —  . 


488  UNIFIED  MATHEMATICS 

7.   Limiting  forms.  —  The  limiting  forms   corresponding  to 
the  ellipsoid  are  given  by  equations  of  the  type 

3.2        y2        22 


(x-h?     (y-ky  +  (inJI2  =  o, 
o2  62  c2 

the  first  of  which  represents  the  point  0  (0,  0,  0)  and  the 
second  the  point  (h,  Jc,  I). 

The  method  of  approaching  this  limit  is  best  indicated  by 
writing  the  equation  as 

o2  +  &2+c2~ 

As  k  approaches  0,  the  semi-axes  Vfca2,  Vfcft2,  and  Vfcc2  ap- 
proach 0  as  a  limit. 

In  a  similar  way  the  hyperboloid  equations  approach,  as 
limits,  the  equations  representing  cones  asymptotic  to  the 
given  hyperboloids. 

The  limiting  forms  of  the  paraboloids  are  equations  in  two 
variables,  and  reduce  to  two  planes,  or  to  cylinders. 

The  limiting  forms  of  equations  in  two  variables  represent- 
ing cylinders  correspond,  with  proper  and  more  or  less  evident 
changes,  to  the  limiting  forms  of  the  corresponding  equations 
in  plane  analytics. 

Thus,  any  equation  of  the  second  degree  f(x,  y,  z)  =  0, 
whether  in  one  or  two  or  three  variables,  of  which  the  left- 
hand  member  can  be  factored  into  two  real  linear  factors  in 
the  variables,  represents  two  planes  which  constitute  also  a 
type  of  quadric  surface. 

8.  Applications.  —  The  applications  of  the  conic  sections 
which  have  been  given  in  plane  analytics  are  strictly  applica- 
tions of  surfaces,  or  solids  having  these  surfaces  as  boundaries. 

Thus,  a  bridge  having  a  parabolic  arch  uses  a  solid  having  a 
parabolic  cylinder  as  bounding  surface. 


SOLID  ANALYTICS:  QUADRIC  SURFACES       489 

The  equation  of  the  paraboloid  in  the  Hill  Auditorium,  with 
the  foot  as  unit  of  length,  is  y2  +  z2  =  70.02  x  ;  the  skylight  in 
the  ceiling  of  the  Hill  Auditorium  is  bounded  by  an  elliptical 
cylinder, 

2>2  y2 

—  +  •£•-  =  1,  dimensions  in  feet. 
762     502 

Any  of  the  automobile  reflectors  are  paraboloids,  in  general,  of 
revolution.  Hyperboloids  are  used  as  revolving  cones  in  the 
manufacture  of  iron  pipes  ;  these  pipes  are  passed  between 
two  revolving  cones  whose  axes  are  inclined  at  90°  to 
straighten  the  pipes. 

9.   Circular  sections.  —  Given  the  ellipsoid  represented  by 


a2 

the   question   arises   as   to   what  planes  cut  this   surface  in 
circular  sections. 

The  method  which  we  have  given  above,  under  section  7, 
for  determining  the  nature  of  the  curve  cut  out  of  the  cone  by 
the  plane  y  =  mx  +  A;  applies  to  this  problem.  In  the  ellipsoid 
above  planes  y  =  k  cut  the  surface  in  ellipses.  The  develop- 
ment as  given  shows  that  any  plane  y  =  mx  cuts  this  surface 
in  a  curve  given  by  the  intersection,  also,  of  the  surface, 


z2_  * 
~  ~    ' 


and  the  given  plane,  or  a  curve, 


=1 


a2(l  +  m2)      &2(1  +  m2)      c2 

(?/   "™~    /Hi'  [    tJ   ^^   0 

and  the  line  f*      A  as  axes  of 
z  =  0  I  x= 0 

reference,  both  lying  in  the  plane  of  the  section.     This  curve 
can  be  written, 

c2(62  -  U2m2)a;'2  +  a262(l  +  m2)z'2  =  aWtl  +  m2). 


490  UNIFIED  MATHEMATICS 

Equating  the  coefficients  of  x'2  and  z'2  gives 


ra  =  ± 


a2(c2  -  62) 
6x/a2-c2 


The  two  planes,  y  =  ±       _      ^  x,  and  all  planes  parallel  to 


them,  cut  this  surface  in  circular  sections.  For  these  to  be  real 
planes  c  must  be  intermediate  in  value  between  a  and  6.  If  c 
is  not  intermediate  between  a  and  6,  then  planes  either  of  the 
form  y  =  raz  or  y  —  nx  will  make  real  circular  sections. 

The  method  applies  to  elliptic  cylinders,  to  elliptic  cones, 
and  to  hyperboloids,  as  well  as  to  the  ellipsoid. 

A  simpler  method,  assuming  c  as  the  intermediate  value,  is  to 

cc2      y2 
find  in  the  ellipse  —  \-  ^-  =  1   a   diameter   of  length  2  c  ;  this 

Ct          0 

diameter  with  the  z-axis  determines  the  plane  of  a  circular 
section. 

10.    Tangent  planes  and  tangent  lines.  —  The  formula 

Ax&  +  By$  +  G(x  +  ajj)  +  F(y  +  7/t)  +  C  =  0, 
which  gives  the  tangent  to 


at  the  point  PI(XI,  yi)  on  the  curve  applies  in  space  analytics, 
with  the  addition  of  the  corresponding  z2  and  z  terms,  to 
give  the  tangent  plane  to  the  quadric  surface  at  a  point 
PI(XI}  yl}  Zi)  on  the  surface. 

The  tangent  plane  at  PI(XI}  y1}  Zj)  to  the  surface 

A&  +  By2  +Cz2+2  Gx  +  2'Fy  +  2Ez  +  K=Q 
is  given  by  the  equation, 

+  G(x 


SOLID  ANALYTICS:  QUADRIC  SURFACES       491 

When  the  point  PI(X^  yl}  2t)  is  not  on  the  quadric  surface, 
this  equation  represents  not  the  tangent  plane  but  the  polar 
plane  of  the  point  (xlf  yly  Zj)  with  respect  to  the  surface.  For 
any  point  outside  of  the  surface,  tangent  planes  to  the  surface 
have  their  points  of  tangency  situated  upon  a  plane,  the  polar 
plane  of  the  point  PI(XI}  yi,  «i).  A  more  complete  discussion  of 
the  polar  plane  would  reveal  many  other  points  of  similarity 
between  the  polar  plane  as  related  to  its  quadric  surface  and  the 
polar  line  as  related  to  its  conic. 

The  intersection  of  a  tangent  plane  at  a  point  PI(XI}  yl}  z^ 
on  the  surface  with  any  other  plane  through  Pt  gives  a  tan- 
gent line  to  the  surface  at  P^ 

11.  Ruled  surfaces.  Generating  lines.  —  Any  surface  which 
can  be  generated  by  the  motion  of  a  straight  line  moving 
according  to  some  law  is  termed  a  ruled  surface.  Evidently, 
by  its  method  of  generation,  such  a  surface  has  straight-line 
elements,  called  rectilinear  generators,  which  lie  wholly  upon 
the  surface. 

Certain  of  the  quadric  surfaces  are  ruled  surfaces.  Evi- 
dently all  the  cylinders,  the  cones,  and  the  pairs  of  planes 
belong  in  this  class.  The  ellipsoid,  being  confined  to  a  finite 
portion  of  space,  does  not  have  right-line  elements  lying 
wholly  upon  the  surface ;  nor  do  the  elliptic  paraboloid  and 
the  hyperboloid  of  two  sheets  have  right-line  elements. 

The  hyperbolic  paraboloid  and  the  hyperboloid  of  one 
sheet  do  have  rectilinear  generators.  We  will  find  the  equa- 
tion of  the  families  of  lines  which  lie  wholly  upon  one  of  the 
surfaces  in  question ;  the  method  will  apply  to  the  other  ruled 
quadric  surfaces. 

Any  point  upon  the  hyperboloid 
o;2  _  172      z2 
a2      62      c2  ~ 
very  evidently  satisfies  the  equation 

*?  _  £  =  1  _  ?- 
a2     62~         c2' 


492 


UNIFIED  MATHEMATICS 


which  may  be  written, 


This  indicates  that  any  point  which  satisfies  the  pair  of  linear 
equations  /        v\ 

-J.tfi-!\ 

a      b        \        c 

x  +  1  = 
a     b 

will  satisfy  the  equation  of  our  surface  since  it  will  make  the 
product  represented  by  the  left-hand  member  of  our  equation, 

18,17,16,15 14  13   12     11     10      9      8       7     6    5  4  3.2 

19.20212223     24     25     26    27      28    293031.32.1 


30,31,32    123 
29,28  27    26      25 


9       10     11,12,13,14 
19      18     17,16,15 


The  right-line  generators  on  this  hyperboloid  of  revolution  are  formed  by 
connecting  corresponding  points  on  two  circular  sections 

An  elliptic  section  is  also  indicated. 

in  the  second  form  above,  equal  to  the  product  of  the  factors, 
representing  the  right-hand  member.      But  every  point  which 


SOLID  ANALYTICS:  QUADRIC  SURFACES      493 

satisfies  the  pair  of  equations  for  any  given  value  of  Tc  lies 
upon  a  straight  line,  the  intersection  of  the  two  planes  given 
by  the  linear  equations.  Hence,  every  point  upon  this  line 
lies  upon  the  given  surface  for  any  value  of  k. 

It  can  be  shown  that  no  two  lines  of  this  family  of  lines, 
i.e.  no  two  lines  given  by  two  values  of  k,  intersect. 

Another  family  of  lines  also  lies  upon  this  surface.  The 
equations  of  this  second  family  of  lines,  with  the  parameter 
fc,  are  as  follows : 


a     b      k 

Every  member  of  this  family  of  lines  can  be  shown  to  inter- 
sect every  member  of  the  preceding  family  and  no  member 
of  its  own  family. 

PROBLEMS 

1.  Find  the  equations  of  the  rectilinear  generators  of  the 
following  surfaces : 

a.   x2  —  y2—   z2  =  0. 

6.    x*  +  y2-    z2=16. 

c.  x2  —  yz  —  4  z  —  0. 

d.  x*-y*  =0. 

2.  Find  the  circular  sections  of  the  following  surfaces : 

™2              ,/2          ,.2 
•  *                      il  6  -\ 

ft  ,        __  jZ ^_  .  _      —      I 


25 

16 

(.) 

b. 

a2 
25 

y2 
16 

z- 
9 

=  0. 

c. 

re2 

+  4  y2 

=  9z. 

d. 

afl 

+  4y2 

=  9. 

e. 

100 

+  36~ 

16 

=  1. 

494  UNIFIED  MATHEMATICS 

3.  Write  the  equations  of  the  tangent  planes  to  each  of  the 
surfaces  in  the  preceding  problem  at  the  point  (a^,  ylt  z^)  in  each 
case  upon  the  given  surface. 

4.  In  problem  1  a,  above,  take  ki  =  1  and  fe2  =  2  and  show- 
that  these  two  lines  of  the  same  family  of  rectilinear  generators 
do  not  intersect.      Write  the  second  family  of  rectilinear  gen- 
erators of  the  same  surface  and  show  that,  taking  k  =  1  (any 
other  value  would  do),  this  line  does  intersect  a  given  line 
(&!  =  1)  of  the  first  set.      How  could  you  make  this  proof 
general ?     • 


TABLES 

PAGE 

Constants  with  their  logarithms • 496 

Squares  and  cubes  of  integers,  1  to  100 497 

Four-place  logarithms  of  numbers,  100  to  999 498 

Four-place  logarithms  of  numbers,  1000  to  1999 500 

Four-place  logarithms  of  sines  and  cosines,  10'  intervals  ....  502 

Four-place  logarithms  of  tangents  and  cotangents,  10'  intervals  .  .  504 
Four-place  logarithms  of  sines,  0°  to  9°,  and  of  cosines,  81°  to  90°,  by 

minutes 506 

Four-place  logarithms  of  tangents,  0°  to  9°,  and  of  cotangents,  81°  to 

90°,  by  minutes • 507 

Four-place  natural  sines  and  cosines,  10'  intervals 508 

Four-place  natural  tangents  and  cotangents,  10'  intervals  ....  510 

Radian  measure  of  angles 612 

Minutes  as  decimals  of  one  degree  ;  &  and  e~z  tables 513 

The  accumulation  of  1  at  the  end  of  n  years 514 

The  present  value  of  1  due  in  n  years 515 

The  accumulation  of  an  annuity  of  1  per  annum  at  the  end  of 

n  years 616 

The  present  value  of  an  annuity  of  1  per  annum  for  n  years  .  .  .  517 
The  annual  sinking  fund  which  will  accumulate  to  1 ,  or,  by  addition 

of  i,  the  annuity  which  1  will  purchase 618 


495 


496 


UNIFIED  MATHEMATICS 


Constants  with  their  logarithms. 

NI-MBER  LOGARITHM 

Base  of  natural  logarithms e  =  2.71828183  0.4342945 

Modulus  of  common  logarithms    .     .     .  u  =  0.43429448  9.0377843-10 

Circumference  of  a  circle  in  degrees  .     .     =  360  2.5563025 

Circumference  of  a  circle  in  minutes      .      =  21600  4.3344538 

Circumference  of  a  circle  in  seconds      .     =1296000  6.1126050 

Radian  expressed  in  degrees     .     .     .     .      =  57.29578  1.7581226 

Radian  expressed  in  minutes    ....     =3437.7468  3.5362739 

Radian  expressed  in  seconds    ....      =206264.806  5.3144251 

Ratio  of  a  circumference  to  diameter     .  TT  =  3.14156592  0.4971499 
IT  =  3.14159265358979323846264338328 

VOLUMES  AXD  WEIGHTS 

Cubic  inches  in  1  gallon  (U   S.)    .    .     .     =231  2.3636 

Gallons  in  1  cubic  foot =  7.48  .8739 

Cubic  inches  in  1  bushel =2150.4  3.3325 

Pounds  per  cubic  foot  of  water  (4°  C.)   .     =62.43  1.79.">1 

Pounds  per  cubic  foot  of  air  (0°  C.)  .     .     =  0.0807  8.9069-10 

Cubic  feet  in  1  cubic  meter       .     .          .     =  35.32  1.5480 

Cubic  meters  in  1  cubic  yard    ....     =0.76  9.8808-10 

Cubic  inches  in  1  liter =  61.03  1.7855 

Liters  in  1  gallon  (U.  S.) =3.786  -5783 

Pounds  in  1  kilogram =  2.2  or  2.205         .3434 

Metric  ton  in  pounds =  2205  3.3434 

Volume  of  sphere,  f  wr* =  4.1888r3 

4.1888  -6221 

LENGTHS  AND  AREAS 

Inches  in  1  meter  (by  Act  of  Congress)       =  39.37  1.5952 
Feet  in  1  rod,  16.5  ;  yards  in  1  rod     .     .     =  5.5 

Square  feet  in  1  acre =  43560  4.6391 

160  square  rods  =  1  acre  ;  640  acres  =  1  square  mile  ; 
3.281  feet  =  1.094  yards  =  1  meter. 


TABLES 


497 


Squares  and  cubes  of  integers,  1  to  100. 
Square  roots  and  cube  roots  of  1  to  100.          Reciprocals  of  1  to  100. 


ss 

D 

« 

D  g 

W  H 

a  o 

%  £o 

• 

t> 

a 

D  ° 

H  H 

n  o 

La 

02 

D 
O 

$x 

&  O 
Utf 

H  o 

or 

02 

O 

yO 

D  O 

Utf 

w" 

n 

n2 

n3 

VH 

V7i 

1 

n 

n2 

n3 

Vn 

\'n 

it 

n 

n 

i 

1 

l 

1.000 

1.000 

1.0000 

51 

2,601 

132,651 

7.141 

3.708 

.0196 

•i 

4 

8 

2.414 

1.260 

.5000 

52 

2,704 

140,608 

7.211 

3.733 

.0192 

I 

9 

27 

1.732 

1.442 

.3333 

53 

2,809 

148,877 

7.280 

3.756 

.0189 

4 

16 

64 

2.000 

1.587 

.2500 

54 

2,916 

157,464 

7.348 

3.780 

.0185 

5 

25 

125 

2.236 

1.710 

.2000 

55 

3,025 

166,375 

7.416 

3.803 

.0182 

6 

36 

216 

2.449 

1.817 

.1667 

56 

3,136 

175,616 

7.483 

3.826 

.0179 

7 

49 

343 

2.646 

1.913 

.1429 

57 

3,249 

185,193 

7.550 

3.849 

.0175 

8 

64 

512 

2.828 

2.000 

.1250 

58 

3,364 

195,112 

7.616 

3.871 

.0172 

9 

81 

729 

3.000 

2.080 

.1111 

59 

3,481 

205,379 

7.681 

3.893 

.0170 

10 

100 

1,000 

3.162 

2.154 

.1000 

60 

3,600 

216,000 

7.746 

3.915 

.0167 

11 

121 

1,331 

3.317 

2.224 

.0909 

61 

3,721 

226,981 

7.810 

3.936 

.0164 

12 

144 

1,728 

3.464 

2.289 

.0833 

62 

3,844 

238,328 

7.874 

3.958 

.0161 

13 

169 

2,197 

3.606 

2.351 

.0769 

63 

3,969 

250,047 

7.937 

3.979 

.0159 

14 

196 

2,744 

3.742 

2.410 

.0714 

64 

4,096 

262,144 

8.000 

4.000 

.0156 

1.5 

225 

3,375 

3.873 

2.466 

.0667 

65 

4,225 

274,625 

8.062 

4.021 

.0154 

1C, 

256 

4,096 

4.000 

2.520 

.0625 

66 

4,356 

287,496 

8.124 

4.041 

.0152 

17 

289 

4,913 

4.123 

2.571 

.0588 

67 

4,489 

300,763 

8.185 

4.062 

.0149 

IS 

324 

5,832 

4.243 

2.621 

.0556 

68 

4,624 

314,432 

8.246 

4.082 

.0147 

lit 

361 

6.859 

4,359 

2.668 

.0526 

69 

4,761 

328,509 

8.307 

4.102 

.0145 

20 

400 

8,000 

4.472 

2.714 

.0500 

70 

4,900 

343,000 

8.367 

4.121 

.0143 

21 

441 

9,261 

4.583 

2.759 

.0476 

71 

5,041 

357,911 

8.426 

4.141 

.0141 

22 

484 

10,648 

4,690 

2.802 

.0455 

72 

5,184 

373,248 

8.485 

4.160 

.0139 

23 

529 

12,167 

4.796 

2.844 

.0436 

73 

5,329 

389,017 

8.544 

4.179 

.0137 

"it 

576 

13,824 

4.899 

2.884 

.0417 

74 

5,476 

405,224 

8.602 

4.198 

.0135 

2.5 

625 

15,625 

5.000 

2.924 

.0400 

75 

5,625 

421,875 

8.660 

4.217 

.0133 

26 

676 

17,576 

5.099 

2.962 

.0385 

76 

5,776 

438,976 

8.718 

4.236 

.0132 

27 

729 

19,683 

5.196 

3.000 

.0370 

77 

5,929 

456,533 

8.775 

4.254 

.0130 

•is 

784 

21,952 

5.292 

3.037 

.0357 

78 

'  6,084 

474,552 

8.832 

4.273 

.0128 

29 

841 

24,389 

5.385 

3.072 

.0345 

79 

6,241 

493,039 

8.888 

4.291 

.0127 

30 

900 

27,000 

5.477 

3.107 

.0333 

80 

6,400 

512,000 

8.944 

4.309 

'.0125 

31 

961 

29,791 

5.568 

3.141 

.0323 

81 

6,561 

531,441 

9.000 

4.327 

.0123 

32 

1,024 

:^,7r,s 

5.657 

3.175 

.0313 

82 

6,724 

551,368 

9.055 

4.344 

.0122 

33 

1,089 

35,937 

5.745 

3.208 

.0303 

83 

6,889 

571,787 

9.110 

4.362 

.0120 

34 

1,156 

39,304 

5.831 

3.240 

.0294 

84 

7,056 

592,704 

9.165 

4.380 

.0119 

3.5 

1,225 

42,875 

5.916 

3.271 

.0286 

85 

7,225 

614,125 

9.220 

4.397 

.0118 

315 

1,296 

46,656 

6.000 

3.302 

.0278 

86 

7,396 

636,056 

9.274 

4.414 

.0116 

37 

1,369 

50,653 

6.083 

3.332 

.0270 

87 

7,569 

658,503 

9.327 

4.431 

.0115 

3S 

1,111 

:>I>7L> 

6.164 

3.362 

.0263 

88 

7,744 

681,472 

9.381 

4.448 

.0114 

39 

1,521 

6.245 

3.391 

.0256 

89 

7,921 

704,969 

9.434 

4.465 

.0112 

40 

1,600 

64,000 

6.325 

3.420 

.0250 

90 

8,100 

729,000 

9.487 

4.481 

.0111 

41 

1,681 

68,921 

6.403 

3.448 

.0244 

91 

8,281 

753,571 

9.539 

4.498 

.0110 

42 

1,764 

74,088 

6.481 

3.476 

.0238 

92 

8,464 

77s.oss 

9.592 

4.514 

.0109 

43 

1,849 

79,507 

6.557 

3.503 

.0233 

93 

8,649 

804,357 

9.644 

4.531 

.0108 

44 

1,930 

85,184 

6.633 

3.530 

.0227 

94 

8,836 

830,584 

9.695 

4.547 

.0106 

4.5 

2,025 

91,125 

6.708 

3.557 

.0222 

95 

9,025 

857,375 

9.747 

4.563 

.0105 

4(i 

2,116 

97,336 

6.782 

3.583 

.0217 

96 

9,216 

884,736 

9.798 

4.579 

.0104 

47 

2,209 

103,823 

6.856 

3.609 

.0213 

97 

9,409 

912,673 

9.849 

4.595 

.0103 

4S 

2,304 

110,592 

6.928 

3.634 

.0208 

98 

9,604 

941,192 

9.899 

4.610 

.0102 

49 

2,401 

117,649 

7.000 

3.659 

.0204 

99 

9,801 

970,299 

9.950 

4.626 

.0101 

.50 

2,500 

125,000 

7.071 

3.684 

.0200 

100 

10,000 

1,000,000 

10.000 

4.642 

.0100 

X 

n* 

n3 

Vn 

^ 

1 

n 

n* 

n» 

Vn 

V^ 

1 

n 

n 

498 


UNI  FIED  MATHEMATICS 


Logarithms  of  numbers  from  100  to  549. 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

To  avoid  inter- 
polation in  the 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

first  ten  lines. 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

following  pages. 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

22 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

1   2.2 
2  4.4 

3  6.6 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

4  8.8 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

5  11.0 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

6  13.2 
7154 

18 

2553 

2577 

2601 

2025 

2648 

2672 

2695 

2718 

2742 

2765 

8  17^6 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

9  19.8 

21  20  19 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

1  2.1  2.0  1.9 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

2  4.2  4.0  3.8 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

3  6.3  6.0  5.7 
4  84  80  76 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

5  10!5  lo!o  9.5 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

6  12.6  12.0  11.4 

7  14.7  14.0  13.3 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

8  16.8  16.0  15.2 
9  18.9  18.0  17.1 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

18  17  16 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

1  1.8  1.7  1.6 
2  36  34  32 

29 

462.4 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

3  5^4  5.1  4.8 

4  7.2  6.8  6.4 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

5  9.0  8.5  8.0 
6  10  8  10  2  9.6 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

8  14.'4  13^6  12.'8 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

9  16.2  15.3  14.4 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

15  14  13 

1   1.5  1.4  1.3 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

2  3.0  2.8  2.6 
3  4.5  4.2  3.9 

4  6.0  5.6  5.2 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

5  7.5  7.0  6.5 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

6  9.0  8.4  7.8 

37 

5682 

5694 

5705 

5717 

'5729 

5740 

5752 

5763 

5775 

5786 

7  10.5  9.8  9.1 
8  12  0  112  104 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

9  13^5  12!6  11.7 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

11     12 

1  1.1    1   1.2 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

2  2.2    2  2.4 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

3  3.3    3  3.6 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

4  4.4    4  4.8 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

5  5.5    5  6.0 
666    6  72 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

7  7.7    7  8.4 

8  8.8    8  9.6 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

9  9.9    9  10.8 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

£  H 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

j?  0 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

3  '2 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

e  3 
11 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

•3 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

1  1 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

1  J 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

*  s?" 

TABLES 


499 


Logarithms  of  numbers  from  550  to  999. 


55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

w 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

,« 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

1 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

•4 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

I 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

1 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

.2" 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

^ 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

i 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

M 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

1 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

10 

3 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

jj 

1 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

I 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

a 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

V 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

9 
a 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

1 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

1 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

3 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

9949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

3 


5 


6 


8 


80 
81 

82 
83 

84 


9031  9036  9042  9047 
9085  9090  9096  9101 
9138  9143  9149  9154 
9191  9196  9201  9206 
9243  9248  9253  9258 


9053  9058 
9106  9112 
9159  9165 
9212  9217 
9263  9269 


9063  9069  9074  9079 
9117  9122  9128  9133 
9170  9175  9180  9186 
9222  9227  9232  9238 
9274  9279  9284  9289 


85 
86 
87 
88 
89 


9294  9299  9304  9309 
9345  9350  9355  9360 
9395  9400  9405  9410 
9445  9450  9455  9460 
9494  9499  9504  9509 


9315  9320 
9365  9370 
9415  9420 
9465  9469 
9513  9518 


9325  9330  9335  9340 
9375  9380  9385  9390 
9425  9430  9435  9440 
9474  9479  9484  9489 
9523  9528  9533  9538 


90 
91 
92 
93 
94 


9542  9547  9552  9557 
9590  9595  9600  9605 
9638  9643  9647  9652 
9685  9689  9694  9699 
9731  9736  9741  9745 


9562  9566 
9609  9614 
9657  9661 
9703  9708 
9750  9754 


9571  9576  9581  9586 
9619  9624  9628  9633 
9666  9671  9675  9680 
9713  9717  9722  9727 
9759  9763  9768  9773 


95 

9« 


9777  9782  9786  9791 
9823  9827  9832  9836 
9868  9872  9877  9881 
9912  9917  9921  9926 
9956  9961  9965  9969 


9795  9800 
9841  9845 
9886  9890 
9930  9934 
9974  9978 


9805  9809  9814  9818 
9850  9854  9859  9863 
9894  9899  9903  9908 
9939  9943  9948  9952 
9983  9987  9991  9996 


G 


8 


500 


UNIFIED  MATHEMATICS 


Logarithms  of  numbers  between  1000  and  1499. 


0 

1 

3 

3 

4 

5 

6 

7 

8 

9 

1M 

0000 

0004 

0009 

0013 

0017 

0022 

0026 

0030 

0035 

0039 

101 

0043 

0048 

0052 

0056 

0060 

0065 

0069 

0073 

0077 

0082 

103 

0086 

0090 

0095 

0099 

0103 

0107 

0111 

0116 

0120 

0124 

103 

0128 

0133 

0137 

0141 

0145 

0149 

0154 

0158 

0162 

0166 

104 

0170 

0175 

0179 

0183 

0187 

0191 

0195 

0199 

0204 

0208 

105 

0212 

0216 

0220 

0224 

0228 

0233 

0237 

0241 

0245 

0249 

106 

0253 

0257 

0261 

0265 

0269 

0273 

0278 

0282 

0286 

0290 

107 

0294 

0298 

0302 

0306 

0310 

0314 

0318 

0322 

0326 

0330 

108 

0334 

0338 

0342 

0346 

0350 

0354 

0358 

0362 

0366 

0370 

109 

0374 

0378 

0382 

0386 

0390 

0394 

0398 

0402 

0406 

0410 

110 

0414 

0418 

0422 

0426 

0430 

0434 

0438 

0441 

0445 

0449 

111 

0453 

0457 

0461 

0465 

0469 

0473 

0477 

0481 

0484 

0488 

113 

0492 

0496 

0500 

0504 

0508 

0512 

0515 

0519 

0523 

0527 

113 

0531 

0535 

0538 

0542 

0546 

0550 

0554 

0558 

0561 

0565 

114 

0569 

0573 

0577 

0580 

0584 

0588 

0592 

0596 

0599 

0603 

115 

0607 

0611 

0615 

0618 

0622 

0626 

0630 

0633 

0637 

0641 

116 

0645 

0648 

0652 

0656 

0660 

0663 

0667 

0671 

0674 

0678 

117 

0682 

0686 

0689 

0693 

0697 

0700 

0704 

0708 

0711 

0715 

118 

0719 

0722 

0726 

0730 

0734 

0737 

0741 

0745 

0748 

0752 

119 

0755 

0759 

0763 

0766 

0770 

0774 

0777 

0781 

0785 

0788 

130 

0792 

0795 

0799 

0803 

0806 

0810 

0813 

0817 

0821 

0824 

131 

0828 

0831 

0835 

0839 

0842 

0846 

0849 

0853 

0856 

0860 

133 

0864 

0867 

0871 

0874 

0878 

0881 

0885 

0888 

0892 

0896 

133 

0899 

0903 

0906 

0910 

0913 

0917 

0920 

0924 

0927 

0931 

134 

0934 

0938 

0941 

0945 

0948 

0952 

0955 

0959 

0962 

0966 

0 

1 

3 

3 

4 

5 

6 

7 

8 

9 

135 

0969 

0973 

0976 

0980 

0983 

0986 

0990 

0993 

0997 

1000 

136 

1004 

1007 

1011 

1014 

1017 

1021 

1024 

1028 

1031 

1035 

137 

1038 

1041 

1045 

1048 

1052 

1055 

1059 

1062 

1065 

1069 

138 

1072 

1075 

1079 

1082 

1086 

1089 

1092 

1096 

1099 

1103 

139 

1106 

1109 

1113 

1116 

1119 

1123 

1126 

1129 

1133 

1136 

130 

1139 

1143 

1146 

1149 

1153 

1156 

1159 

1163 

1166 

1169 

131 

1173 

1176 

1179 

1183 

1186 

1189 

1193 

1196 

1199 

1202 

133 

1206 

1209 

1212 

1216 

1219 

1222 

1225 

1229 

1232 

1235 

133 

1239 

1242 

1245 

1248 

1252 

1255 

1258 

1261 

1265 

1268 

134 

1271 

1274 

1278 

1281 

1284 

1287 

1290 

1294 

1297 

1300 

135 

1303 

1307 

1310 

1313 

1316 

1319 

1323 

1326 

1329 

1332 

136 

1335 

1339 

1342 

1345 

1348 

1351 

1355 

1358 

1361 

1364 

137 

1367 

1370 

1374 

1377 

1380 

1383 

1386 

1389 

1392 

1396 

138 

1399 

1402 

1405 

1408 

1411 

1414 

1418 

1421 

1424 

1427 

139 

1430 

1433 

1436 

1440 

1443 

1446 

144'9 

1452 

1455 

1458 

140 

1461 

1464 

1467 

1471 

1474 

1477 

1480 

1483 

1486 

1489 

141 

1492 

1495 

1498 

1501 

1504 

1508 

1511 

1514 

1517 

1520 

143 

1523 

1526 

1529 

1532 

1535 

1538 

1541 

1544 

1547 

1550 

143 

1553 

1556 

1559 

1562 

1565 

1569 

1572 

1575 

1578 

1581 

144 

1584 

1587 

1590 

1593 

1596 

1599 

1602 

1605 

1608 

1611 

145 

1614 

1617 

1620 

1623 

1626 

1629 

1632 

1635 

1638 

1641 

146 

1644 

1647 

1649 

1652 

1655 

1658 

1661 

1664 

1667 

1670 

147 

1673 

1676 

1679 

1682 

1685 

1688 

1691 

1694 

1697 

1700 

148 

1703 

1706 

1708 

1711 

1714 

1717 

1720 

1723 

1726 

1729 

149 

1732 

1735 

1738 

1741 

1744 

1746 

1749 

1752 

1755 

1758 

0 

1 

3 

3 

4 

5 

6 

7 

8 

9 

TABLES 


501 


Logarithms  of  numbers  between  1500  and  1999. 


0 

1 

2 

3 

4 

5 

6 

7 

8 

9  . 

150 

1761 

1764 

1767 

1770 

1772 

1775 

1778 

1781 

1784 

1787 

151 

1790 

1793 

1796 

1798 

1801 

1804 

1807 

1810 

1813 

1816 

159 

1818 

1821 

1824 

1827 

1830 

1833 

1836 

1838 

1841 

1844 

153 

1847 

1850 

1853 

1855 

1858 

1861 

1864 

1867 

1870 

1872 

154 

1875 

1878 

1881 

1884 

1886 

1889 

1892 

1895 

1898 

1901 

155 

1903 

1906 

1909 

1912 

1915 

1917 

1920 

1923 

1926 

1928 

156 

1931 

1934 

1937 

1940 

1942 

1945 

1948 

1951 

1953 

1956 

157 

1959 

1962 

1965 

1967 

1970 

J973 

1976 

1978 

1981 

1984 

158 

1987 

1989 

1992 

1995 

1998 

2000 

2003 

2006 

2009 

2011 

159 

2014 

2017 

2019 

2022 

2025 

2028 

2030 

2033 

2036 

2038 

160 

2041 

2044 

2047 

2049 

2052 

2055 

2057 

2060 

2063 

2066 

161 

2068 

2071 

2074 

2076 

2079 

2082 

2084 

2087 

2090 

2092 

162 

2095 

2098 

2101 

2103 

2106 

2109 

2111 

2114 

2117 

2119 

163 

2122 

2125 

2127 

2130 

2133 

2135 

2138 

2140 

2143 

2146 

164 

2148 

2151 

2154 

2156 

2159 

2162 

2164 

2167 

2170 

2172 

165 

2175 

2177 

2180 

2183 

2185 

2188 

2191 

2193 

2196 

2198 

166 

2201 

2204 

2206 

2209 

2212 

2214 

2217 

2219 

2222 

2225 

167 

2227 

2230 

2232 

2235 

2238 

2240 

2243 

2245 

2248 

2251 

168 

2253 

2256 

2258 

2261 

2263 

2266 

2269 

2271 

2274 

2276 

169 

2279 

2281 

2284 

2287 

2289 

2292 

2294 

2297 

2299 

2302 

170 

2304 

2307 

2310 

2312 

2315 

2317 

2320 

2322 

2325 

2327 

171 

2330 

2333 

2335 

2338 

2340 

2343 

2345 

2348 

2350 

2353 

173 

2355 

2358 

2360 

2363 

2365 

2368 

2370 

2373 

2375 

2378 

173 

2380 

2383 

2385 

2388 

2390 

2393 

2395 

2398 

2400 

2403 

174 

2405 

2408 

2410 

2413 

2415 

2418 

2420 

2423 

2425 

2428 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

175 

2430 

2433 

2435 

2438 

2440 

2443 

2445 

2448 

2450 

2453 

176 

2455 

2458 

2460 

2463 

2465 

2467 

2470 

2472 

2475 

2477 

177 

2480 

2482 

2485 

2487 

2490 

2492 

2494 

2497 

2499 

2502 

178 

2504 

2507 

2509 

2512 

2514 

2516 

2519 

2521 

2524 

2526 

179 

2529 

2531 

2533 

2536 

2538 

2541 

2543 

2545 

2548 

2550 

180 

2553 

2555 

2558 

2560 

2562 

2565 

2567 

2570 

2572 

2574 

181 

2577 

257!) 

2582 

2584 

2586 

2589 

2591 

2594 

2596 

2598 

182 

2601 

2603 

2605 

2608 

2610 

2613 

2615 

2617 

2620 

2622 

183 

2625 

2627 

2629 

2632 

2634 

2636 

2639 

2641 

2643 

2646 

184 

2648 

2651 

2653 

2655 

2658 

2660 

2662 

2665 

2667 

2669 

185 

2672 

2674 

2676 

2679 

2681 

2683 

2686 

2688 

2690 

2693 

186 

2695 

2697 

2700 

2702 

2704 

2707 

2709 

2711 

2714 

2716 

187 

2718 

2721 

2723 

2725 

2728 

2730 

2732 

2735 

2737 

2739 

188 

2742 

2744 

2746 

2749 

2751 

2753 

2755 

2758 

2760 

2762 

189 

2765 

2767 

2769 

2772 

2774 

2776 

2778 

2781 

2783 

2785 

190 

2788 

2790 

2792 

2794 

2797 

2799 

2801 

2804 

2806 

2808 

191 

2810 

2813 

2815 

2817 

2819 

2822 

2824 

2826 

L'S2S 

2831 

IK 

2833 

2835 

2838 

2840 

2842 

2844 

2847 

2849 

2851 

2853 

193 

2856 

2858 

2860 

2862 

2865 

2867 

2869 

2871 

2874 

2876 

194 

2878 

2880 

2882 

2885 

2887 

2889 

2891 

2894 

2896 

2898 

195 

2900 

2903 

2905 

2907 

2909 

2911 

2914 

2916 

2918 

2920 

196 

2923 

2925 

2927 

2929 

2931 

2934 

2936 

2938 

2940 

2942 

197 

2945 

2947 

2949 

2951 

2953 

2956 

2958 

2960 

2983 

2964 

198 

2967 

2969 

2971 

2973 

2975 

2978 

2980 

2982 

2984 

2986 

199 

2989 

2991 

2993 

2995 

2997 

2999 

3002 

3004 

3006 

3008 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

502 


UNIFIED  MATHEMATICS 


Log  sin  A°  from  0°  to  45  °- 


10'       20' 


30' 


40' 


50'        60'      A°    d. 


0 

7.4637  7648 

9408 

*0658 

*1627 

*2419 

89° 

Do  not  interpolate,  but 

1  8.2419 
2  5428 

3088  3668 
5776  6097 

4179 
6397 

4637 
6677 

5050 
6940 

5428 
7188 

88° 
87° 

use  the  special  table 

3  7188 

7423  7645 

7857 

8059 

8251 

8436 

86° 

which 

gives  these  values 

4  8436 

8613  8783 

8946 

9104 

9256 

9403 

85° 

by  minutes. 

5  9403 

9545  9682 

9816 

9945 

*0070 

*0192 

84° 

6  9.0192 

0311  0426 

0539 

0648 

0755 

0859 

83° 

7  0859 

0961  1060 

1157 

1252 

1345 

1436 

82° 

91 

8   1436 

1525  1612 

1697 

1781 

1863 

1943 

81° 

SO 

9  1943 

2022  2100 

'2176 

2251 

2324 

2397 

80° 

73 

P.  P. 

10  9.2397 

2468  2538 

2606 

2674 

2740 

2806 

79° 

68 

11  2806 

2870  2934 

2997 

3058 

3119 

3179 

78° 

62 

92 

90  88 

86  84 

12  3179 

3238  3296 

3353 

3410 

3466 

3521 

77° 

57 

1 

9.2 

9.0  8.8 

8.6  8.4 

13  3521 

3575  3629 

3682 

3734 

3786 

3837 

76° 

53 

2 

18.4 

18.0  17.6 

17.2  16.8 

14  3837 

3887  3937 

3986 

4035 

4083 

4130 

75° 

49 

3 
4 

27.6 
36.8 

27.0  26.4 
36.0  35.2 

25.8  25.2 
34.4  33.6 

5 

46.0 

45.0  44.0 

43.0  42.0 

15  4130 

4177  4223 

4269 

4314 

4359 

4403 

74° 

46 

6 

55.2 

54.0  52.8 

51.6  50.4 

16  4403 
17   4659 

4447  4491 
4700  4741 

4533 

4781 

4576 
4821 

4618 
4861 

4659 
4900 

73° 

72° 

43 
40 

7 
8 

9 

64.4 
73.6 
82.8 

63.0  61.6 
72.0  70.4 
81.0  79.2 

60.2  58.8 
68.8  67.2 
77.4  75.6 

18  4900 

4939  4977 

5015 

5052 

5090 

5126 

71° 

38 

19  5126 

5163  5199 

5235 

5270 

5306 

5341 

70° 

36 

82 

80   78 

76   74 

1 

8.2 

8.0  7.8 

7.6  7.4 

2 

16.4 

16.0  15.6 

15.2  14.8 

20  9.5341 

5375  5409 

5443 

5477 

5510 

5543 

69° 

34 

3 

24.6 

24.0  23.4 

22.8  22.2 

21  5543 
22  5736 
23  5919 

5576  5609 
5767  5798 
5948  5978 

5641 
5828 
6007 

5673 
5859 
6036 

5704 
5889 
6065 

5736 
5919 
6093 

68° 
67° 
66° 

32 
31 

29 

4 

5 
6 
7 

32.8 
41.0 
49.2 
57.4 

32.0  31.2 
40.0  39.0 
48.0  46.8 
56.0  54.6 

30.4  29.6 
38.0  37.0 
45.6  44.4 
53.2  51.8 

24  6093 

6121  6149 

6177 

6205 

6232 

6259 

65° 

28 

8 

65.6 

64.0  62.4 

60.8  59.2 

9 

73.8 

72.0  70.2 

68.4  66.6 

30' 

72 

70  68 

66   64 

1 

7.2 

7.0  6.8 

6.6  6.4 

25  6259 
26  6418 
27   6570 

6286  6313 
6444  6470 
6595  6620 

6340 
6495 
6644 

6366 
6521 
6668 

6392 
6546 
6692 

6418 
6570 
6716 

64° 
63° 

62° 

27 
25 

24 

2 
3 

4 
5 

14.4 
21.6 
28.8 
36.0 

14.0  13.6 
21.0  20.4 
28.0  27.2 
35.0  34.0 

13.2  12.8 
19.8  19.2 
26.4  25.6 
33.0  32.0 

28  6716 

6740  6763 

6787 

6810 

6833 

6856 

61° 

23 

6 

43.2 

42.0  40.8 

39.6  38.4 

29  6856 

6878  6901 

6923 

6946 

6968 

6990 

60° 

22 

7 
8 

50.4 
57.6 

49.0  47.6 
56.0  54.4 

46.2  44.8 
52.8  51.2 

9 

64.8 

63.0  61.2 

59.4  57.6 

30  9.6990 
31  7118 
32  7242 
33  7361 
34  7476 

7012  7033 
7139  7160 
7262  7282 
7380  7400 
7494  7513 

7055 
7181 
7302 
7419 
7531 

7076 
7201 
7322 
7438 
7550 

7097 
7222 
7342 
7457 
7568 

7118 
7242 
7361 
7476 
7586 

59° 

58° 
57° 
56° 
55° 

21 
21 
20 

18 

1 
2 
3 
4 
5 

62 

6.2 
12.4 
18.6 
24.8 
31.0 

60  58 

6.0  5.8 
12.0  11.6 
18.0  17.4 
24.0  23.2 
30.0  29.0 

56  54 

5.6  5.4 
11.2  10.8 
16.8  16.2 
22.4  21.6 
28.0  27.0 

35  7586 
36  7692 
37   7795 
38  7893 
39  7989 

7604  7622 
7710  7727 
7811  7828 
7910  7926 
8004  8020 

7640 
7744 
7844 
7941 
8035 

7657 
7761 
7861 
7957 
8050 

7778 
7877 
7973 
8066 

7692 
7795 
7893 
7989 
8081 

54° 
53° 
52° 
51° 
50° 

18 

17 

id 

16 

15  . 

6 
7 
8 
9 

1 

37.2 
43.4 
49.6 
55.8 

52 

5.2 

36.0  34.8 
42.0  40.6 
48.0  46.4 
54.0  52.2 

50  48 

5.0  4.8 

33.6  32.4 
39.2  37.8 
44.8  43.2 
50.4  48.6 

46  44 

4.6  4.4 

2 

10.4 

10.0  9.6 

9.2  8.8 

40  9.8081 
41  8169 
42  8255 
43  8338 
44  8418 

8096  8111 
8184  8198 
8269  8283 
8351  8365 
8431  8444 

8125 
8213 
8297 
8378 
8457 

8140 
8227 
8311 
8391 
8469 

8155 
8241 
8324 
8405 
8482 

8169 
8255 
8338 
8418 
8495 

49° 

48° 
47° 
46° 

45° 

15 
14 
14 
13 
13 

3 
4 
5 
f. 
7 
8 
9 

15.6 
20.8 
26.0 
31.2 
36.4 
41.6 
46.8 

15.0  14.4 
20.0  19.2 
25.0  24.0 
30.0  28.8 
35.0  33.6 
40.0  38.4 
45.0  43.2 

13.8  13.2 
18.4  17.6 
23.0  22.0 
27.6  26.4 
32.2  30.8 
36.8  35.2 
41.4  39.6 

60' 


50'      40' 


30' 


20'       10' 


A°   d. 


P.P. 


Log  cos  A°  from  45°  to  90°. 


TABLES 


503 


Log  sin  A°  from  45°  to  90°. 


p.  p. 

A° 

0' 

10' 

207 

30' 

40' 

50' 

60' 

A" 

(1 

p.p. 

45 

8495 

8507 

8520 

8532 

8545 

8557 

8569 

44 

12 

12 

46 

8569 

8582 

8594 

8606 

8618 

8629 

8641 

43 

12 

42  40  38  36 

47 

8641 

8653 

8665 

8676 

8688 

8699 

8711 

12 

2  2^4 

1  4.2  4.0  3.8  3.6 

48 

8711 

8722 

8733 

8745 

8756 

8767 

8778 

41 

11 

3  3.6 

2  84  8.0  7.6  7.2 

49 

8778 

8789 

8800 

8810 

8821 

8832 

8843 

40 

1  1 

4  4.8 

3  12.6  12.0  11.4  10.8 

5  6.0 

4  16.8  16.0  15.2  14.4 

6  7.2 

521.020.0  19.0  18.0 

501 

(.8843 

8853 

8864 

8874 

8884 

8895 

8905 

39 

10 

7  8.4 

625.2  24.022.821.6 

51 

8905 

8915 

8925 

8935 

8945 

8955 

8965 

10 

8  9.6 

729.428.026.625.2 
833.632.030.428.8 
937.836.034.2  32.4 

52 
53 
54 

8965 
9023 
9080 

8975 
9033 
9089 

8985 
9042 
9098 

8995 
9052 
9107 

9004 
9061 
9116 

9014 
9070 
9125 

9023 
9080 
9134 

I 

9 

10 
9 

9  10.8 

34  32  30  28 

55 

56 

9134 
9186 

9142 
9194 

9151 
9203 

9160 
9211 

9169 
9219 

9177 
9228 

9186 

9236 

34 
33 

9 
8 

1  3.4  3.2  3.0  2.8 
2  6.8  6.4  6.0  5.6 
3  10.2  9.6  9.0  8.4 
4  13.6  12.8  12.011.2 

57 

58 
59 

9236 
9284 
9331 

9244 
9292 
9338 

9252 
9300 
9346 

9260 
9308 
9353 

9268 
9315 
9361 

9276 
9323 
9368 

9284 
9331 
9375 

32 
31 

30 

8 
8 

7 

5  17.0  16.0  15.014.0 

6  20.4  19.2  18.0  16.8 

7  23.822.4  21.0  19.6 

601 

t.9375 

8393 

9390 

9397 

9404 

9411 

9418 

''!( 

7 

8  27.2  25.6  24.0  22.4 

61 

9418 

9425 

9432 

9439 

9446 

9453 

9459 

H 

6 

930.628.827.025.2 

62 

9459 

9466 

9473 

9479 

9486 

9492 

9499 

7 

63 

9499 

9505 

9512 

9518 

9524 

9530 

9537 

'Hi 

7 

64 

9537 

9543 

9549 

9555 

9561 

9567 

9573 

M 

6 

• 

26  24  22  20 

3 

1  2.6  2.4  2.2  2.0 

65 

9573 

9579 

9584 

9590 

9596 

9602 

9607 

•J4 

5 

5 

2  5.2  4.8  4.4  4.0 

6G 

9607 

9613 

9618 

9624 

9629 

9635 

9640 

23 

5 

a 

3  7.8  7.2  6.6  6.0 

67 

9640 

9646 

9651 

9656 

9661 

9667 

9672 

•>2 

5 

.3 

4  10.4  9.6  8.8  8.0 
5  13.0  12.0  11.0  10.0 
6  15.6  14.4  13.2  12.0 

68 
69 

9672 
9702 

9677 
9706 

9682 
9711 

9687 
9716 

9692 
9721 

9697 
9725 

9702 
9730 

:!i 

5 
5 

+* 

1 

7  18.2  16.8  15.4  14.0 

•s 

820.8  19.2  17.6  16.0 

30' 

- 

923.421.6  19.8  18.0 

a 

70  9.9730 

9734 

9739 

9743 

9748 

9752 

9757 

19 

5 

71 

9757 

9761 

9765 

9770 

9774 

9778 

9782 

IS 

4 

1 

19  18  17  16 

72 

9782 

9786 

9790 

9794 

9798 

9802 

9806 

17 

4 

y 

1  19  18  17  16 

73 

9806 

9810 

9814 

9817 

9821 

9825 

9828 

it; 

3 

2  3.8  3.6  3.4  3.2 

74 

9828 

9832 

9836 

9839 

9843 

9846 

9849 

15 

3 

1 

3  5.7  5.4  5.1  4.8 

A 

4  76  72  68  6.4 

5  9^5  9.0  8.5  8.0 

75 

9849 

9853 

9856 

9859 

9863 

9866 

9869 

14 

3 

*3 

6  11.4  10.8  10.2  9.6 

76 

9869 

9872 

9875 

9878 

9881 

9884 

9887 

13 

3 

B 

7  13.3  12.6  11.9  11.2 

77 

9887 

9890 

9893 

9896 

9899 

9901 

9904 

12 

3 

8  15.2  14.4  13.6  12.8 
9  17.1  16.2  15.3  14.4 

78 
79 

9904 
9919 

9907 
9922 

9909 
9924 

9912 
9927 

9914 
9929 

9917 
9931 

9919 
9934 

11 

10 

2 
3 

a 

3 

80  9.9934 

9936 

993S 

9940 

9942 

9944 

9946 

,, 

2 

1 

15  14  13  12 

81 

9946 

9948 

9950 

9952 

9954 

9956 

9958 

8 

2 

1  1.5  1.4  1.3  1.2 

82 

OOfiS 

0060 

9961 

9963 

9964 

9966 

9908 

7 

2 

a 

2  3.0  2.8  2.6  2.4 
3  4.5  4.2  3.9  3.6 
4  6.0  5.6  5.2  4.8 

83 
84 

0068 

9976 

9969 
9977 

9971 
9979 

9972 
9980 

9973 
9981 

0078 

9982 

9976 
9983 

1 

5 

1 
1 

5  7.5  7.0  6.5  6.0 

6  9.0  8.4  7.8  7.2 
7  10.5  9.8  9.1  8.4 
8120112104  96 

85 
86 

9983 
9989 

9985 
9990 

9986 

<MMM 

9987 
9992 

9988 
9993 

9989 
9993 

9989 

'.I'.HIl 

4 
1 

0 

1 

9  13^5  12^6  117  10.8 

87 

0004 

9995 

9995 

0006 

0906 

9997 

9997 

1 

0 

88 

9997 

<MM)S 

9998 

0900 

9999 

9999 

'.MMMI 

1 

0 

89 

9999  *0000  *0000 

*0000 

*0000  *0000  *0000 

0 

0 

p.p. 


60'       50'       40' 


30' 


20'      10'      V       A°   d. 


Log  cos  A°  from  0°  to  45°. 


504 


Logarithms  of  tangents  and  cotangents,  0  to  23°. 

10'      20'       30'        40'       50'       60'  A°    d. 


O'tan      7.4637  7648 
log  cot      2.5363  2352 

9409  *0658  *1627  *2419  cot 
0591  *9342  *8373  *7581  tan 

89° 
log 

1°  tan  8.2419 

3089  3669 

4181 

4638 

5053 

5431  cot 

88° 

log  cot  1.7581 

6911  6331 

5819 

5362 

4947 

4569  tan 

log 

Do  not  interpolate, 

2°  tan  8.5431 

5779  6101 

6401 

6682 

6945 

7194  cot 

87° 

but  use  the  special 

log  cot  1.4569 

4221  3899 

3599 

3318 

3055 

2806  tan 

log 

table  for  log  tan  from 

3°  tan  8.7194 

7429  7652 

7865 

8067 

8261 

8446  cot 

86° 

0°  to  9°,  and  log  cot 

log  cot  1.2806 

2571  2348 

2135 

1933 

1739 

1554  tan 

log 

from  81°  to  90°. 

4°  tan  8.8446 

8624  8795 

8960 

9118 

9272 

9420  cot 

85° 

log  cot  1.1554 

1376  1205 

1040 

0882 

0728 

0580  tan 

log 

5°  tan  8.9420 

9563  9701 

9836 

9966*0093*02  16  cot 

84° 

P.P. 

log  cot  1.0580 
6°  tan  9.0216 
log  cot  0.9784 
7°  tan  9.0891 

0437  0299 
0336  0453 
9664  9547 
0995  1096 

0164 
0567 
9433 
1194 

0034  *9907  *9784  tan 
0678  0786  0891  cot 
9322  9214  9109  tan 
1291  1385  1478  cot 

log 

83° 
log 

82° 

82  80  78  76  74 

1  8.2  8.0  7.8  7.6  7.4 
216.4  16.015.615.2  14.8 
324.624.023.422.822.2 

log  cot  0.9109 

9005  8904 

8806 

8709 

8615 

8522  tan 

log 

432.832.031.230.429.6 

541.040  039.038.037.0 

649.248.046.845.644.4 

8°  tan  9.  1478 

1569  1658 

1745 

1831 

1915 

1997  cot 

81° 

87 

.757.456.054.653.251.8 

log  cot  0.8522 
9°  tan  9.  1997 

8431  8342 
2078  2158 

8255 
2236 

8169 
2313 

8085 
2389 

8003  tan 
2463  cot 

log 

HO  ° 

87  865.664.062.460.859.2 
7g  973.872.070.268.466.6 

log  cot  0.8003 

7922  7842 

7764 

7687 

7611 

7537  tan 

log 

78 

72  70  68  66  64 

1  7.2  7.0  6.8  6.6  6.4 

30' 

214.4  14.013.613.212.8 

321.621.020.4  19.819.2 

428.828.027.226.425.6 

10°  tan  9.2463 
log  cot  0.7537 
11°  tan  9.2887 

2536  2609 
7464  7391 
2953  3020 

2680 
7320 
3085 

2750 
7250 
3149 

2819 
7181 
3212 

2887  cot  79° 
71  13  tan  log 
3275  cot  78° 

71 
71 
U 

5  36.0  35.0  34.0  33.0  32.0 
643.242.040.839.638.4 
750.449.047.646.244.8 
8  57  6  56  0  54  4  52  8  51  2 

log  cot  0.7113 

7047  6980 

6915 

6851 

6788 

6725  tan 

log 

66 

9  64.8  63X)  6L2  59^4  57^6 

12°  tan  9.3275 

3336  3397 

3458 

3517 

3576 

3634  cot 

77° 

80 

log  cot  0.6725 

6664  6603 

6542 

6483 

6424 

6366  tan 

log 

80 

62  60  58  56  54 

1  6.2  6.0  5.8  5.6  5.4 

212.4  12.011.611.2  10.8 

13°  tan  9.3634 
log  cot  0.6366 
14°  tan  9.3968 

3691  3748 
6309  6252 
4021  4074 

3804 
6196 
4127 

3859 
6141 
4178 

3914 
6086 
4230 

3968  cot 
6032  tan 
4281  cot 

76° 

log 
75 

M 

66 

.52 

318.618.017.4  16.816.2 
424.824.023.222.421.6 
531.030.029.028.027.0 
6  37  2  36  0  34  8  33  6  32  4 

log  cot  0.6032 

5979  5926 

5873 

5822 

5770 

5719  tan 

log 

.52 

743.442.040.639.237.8 

849.648.046.444.843.2 

15°  tan  9.4281 

4331  4381 

4430 

4479 

4527 

4575  cot 

74° 

49 

9  55.8  54.0  52.2  50.4  48.6 

log  cot  0.5719 

5669  5619 

5570 

5521 

5473 

5425  tan 

log 

49 

53  52  51  50  49 

16°  tan  9.4575 
log  cot  0.5425 
17°  tan  9.4853 

4622  4669 
5378  5331 
4898  4943 

4716 
5284 
4987 

4762 
5238 
5031 

4808 
5192 
5075 

4853  cot 
5147  tan 
51  18  cot 

73° 

log 
72° 

46  1  5.3  5.2  5.1  5.0  4.» 
4fi  2  10.6  10.4  10.2  10.0  9.8 
TV  3  15.9  15.6  15.3  15.0  14.7 
**  4  21.  2  20.8  20.4  20.0  19.6 

log  cot  0.5147 

5102  5057 

5013 

4969 

4925 

4882  tan 

log 

44 

526.526.025.525.024.5 

631.831.230.630.029.4 

7  37  1  36  4  35  7  35  0  34  3 

18°  tan  9.51  18 

5161  5203 

5245 

5287 

5329 

5370  cot 

71° 

42 

8  42  A  4  1'.6  40.8  40.0  39.  2 

log  cot  0.4882 

4839  4797 

4755 

4713 

4671 

4630  tan 

log 

42 

947.746.845.945.044.1 

19°  tan  9.5370 

5411  5451 

5491 

5531 

5571 

5611  cot 

70° 

40 

lx   17   li:   4  \   1-1 

log  cot  0.4630 

4589  4549 

4509 

4469 

4429 

4389  tan 

log 

40 

4O   *«   **>   *w   ** 

1  4.8  4.7  4.6  4.5  4.4 

2  9.6  9.4  9.2  9.0  8.8 

30°  tan  9.5611 

5650  5689 

5727 

5766 

5804 

5842  cot 

69° 

M 

314.4  14.1  13.813.513.2 

log  cot  0.4.389 

4350  4311 

4273 

4234 

4196 

4158  tan 

log 

39 

419.2  18.818.418.017.6 

21°  tan  9.5842 
log  cot  0.4158 

5879  5917 
4121  4083 

5954 
4046 

5991 
4009 

6028 
3972 

6064  cot 
3936  tan 

68° 
log 

37 
37 

524.023.523.022.522.0 
628.828.227.627.026.4 
733.632.932.231.530.8 

22°  tan  9.6064 

6100  6136 

6172 

6208 

6243 

6279  cot 

67° 

36 

838.437.636.836.035.2 

log  cot  0.3936 

3900  3864 

3828 

3792 

3757 

3721  tan 

log 

38 

943.242.341.440.539.6 

60' 


50'     40'       30' 


10' 


P.P. 


Logarithms  of  tangents  and  cotangents,  67°  to  90°. 


TABLES 


505 


Logarithms  of  tangents  and  cotangents,  23°  to  46°. 


P.P. 


10'      20'       30'      40'     50'      60' 


A"    d. 


P.P. 


23°  tan  9.6279  6314  6348  6383  6417  6452  6486  cot  66°  34 
log  cot  0.3721  3686  3652  3617  3583  3548  3514  tan  log  34 
tA"  ton  Q  ,:  1^1;  R^on  RX.HI  RX.V7  R«on  r.i;-..i  ««W7  ,.«t  «£*  •« 


43 

1    4.3 

42 

4.2 

it    tan  ».o-*»o  oozu  000,5 
log  cot  0.3514  3480  3447 

DOS/ 

3413 

OIKU    OOO-l 

3380  3346 

DOS/  COl  DO    .5.5 

33  13  tan  log  33 

1 

33 

3.3 

32 

3.2 

286 

8  4 

2 

6.6 

6.4 

3  12.9 

12.6 

25°  tan  9.  6687  6720 

6752 

6785 

6817  6850 

6882  cot  64°  33 

i 

9.9 

9.6 

4  17.2 

16.8 

log  cot  0.3313  3280 

3248 

3215 

3183  3150 

31  18  tan  log  33 

i 

13.2 

12.8 

5  21.5 

21.0 

26°  tan  9.6882  6914 

6946 

6977 

7009  7040 

7072  cot  63s  32 

."i 

16.5 

16.0 

6  25.8 
7  30  1 

25.2 
29  4 

log  cot  0.3118  3086  3054 

3023 

2991  2960 

2928  tan  log  32 

7 

19.8 
23  1 

19.2 
22  4 

8  34  4 

33  6 

27°  tan  9.  7072  7103 

7134 

7165 

7196  7226 

7257  cot  62°  31 

8 

26.4 

25.6 

938.7 

37.8 

log  cot  0.2928  2897 

2866 

2835 

2804  2774 

2743  tan  log  31 

1 

29.7 

28.8 

28°  tan  9.7257  7287 

7317 

7348 

7378  7408 

7438  cot  61°  30 

log  cot  0.2743  2713 

2683 

2652 

2622  2592 

2562  tan  log  30 

31 

30 

41 

40 

29°  tan  9.7438  7467 

7497 

7526 

7556  7585 

7614  cot  60°  20 

31 

0     1. 

1    4.1 

282 

4.0 

80 

log  cot  0.2562  2533 

2503 

2474 

2444  2415  2386  tan  log  29 

1 
2 

.1 

6.2 

o.u 
6.0 

3  12.3 
4  16.4 

12.0 
16.0 

30°  tan  9.7614  7644 

7673 

7701 

7730  7759 

7788  cot  59°  29 

i 

4 

9.3 

12.4 

9.0 
12.0 

520.5 

20.0 

log  cot  0.2386  2356 

2327 

2299 

2270  2241 

2212  tan  log  29 

B 
i, 

15.5 
18  6 

15.0 
18  0 

6  ?4.6 
7  ?8  7 

24.0 

28  0 

31°  tan  9.  7788  7816 

7845 

7873 

7902  7930 

79.>  cot  58°  28 

7 

21.7 

21.0 

8  32  8 

32  0 

log  cot  0.2212  2184 

2155 

2127 

2098  2070 

2042  tan  log  28 

8 

24.8 

24.0 

9MJ 

36.0 

32°  tan  9.  7958  7986  8014 

8042 

8070  8097 

8125  cot  57   28 

9  27.9 

27.0 

log  cot  0.2042  2014 

1986 

1958 

1930  1903 

1875  tan  log  28 

39 

38 

33°  tan  9.  81  25  8153 

8180 

8208 

8235  8263 

8290  cot  56°  27 

29 

28 

1    3.9 
2    78 

3.8 
7  6 

log  cot  0.1875  1847 
34°  tan  9.8290  8317 

1820 
8344 

1792 
8371 

1765  1737  1710  tan  log  27 
8398  8425  8452  cot  55°  27 

1 

2 

2.9 
5.8 

2.8 
5.6 

3  11> 
4  15.6 
5  19  5 

11  A 
11.4 
19  0 

log  cot  0.1710  1683 

1656 

1629 

1602  1575 

1548  tan  log  27 

I 

4 
5 

8.7 
11.6 
14.5 

8.4 
11.2 
14.0 

i>  2.:5  A 
7  27.3 
8312 

22^8 
26.6 
304 

35°  tan  9.8452  8479 

8506 

30' 

8533 

8*559  8586 

861  3  cot  54°  27 

6  17.4 
7  20.3 

8  2:5.2 

16.8 
19.6 
22.4 

9  35.1 

34.2 

log  cot  0.1548  1521 
36°  tan  9.S613  8639 

1494 

8666 

1467 
8692 

1441  1414 

8718  8745 

1387  tan  log  27 
8771  cot  53°  26 

9  26.1 

25.2 

log  cot  0.1387  1361 

1334 

1308 

1282  1255 

1229  tan  log  26 

37 

36 

37°  tan  9.8771  8797 

8824 

8850 

8876  8902 

8928  cot  52°  26 

27 

26 

1    3.7 

2    7.4 

3.6 
7.2 

log  cot  0.1229  1203 

1176 

1150 

1124  1098 

1072  tan  log  26 

1 

1 

2.7 
5.4 

2.6 
5.2 

3  11.1 

10.8 

1 

8.1 

7.8 

4  14.8 

14.4 

38°  tan  9.8928  8954 

8980 

9006 

9032  9058 

9084  cot  51°  26 

I 

10.8 

10.4 

5  18.5 

r,  -22  -2 

7  25  9 

18.0 
21.6 
25  2 

log  cot  0.1072  1046 
39°  tan  9.9084  9110 

1020 
9135 

0994 
9161 

0968  0942 
9187  9212 

091  6  tan  log  2(i 
9238  cot  50s  26 

5 

a 

7 

13.5 
16.2 

IS  'I 

13.0 
15.6 

18  2 

28^8 

log  cot  0.0916  0890  0865 

0839 

0813  0788 

0762  tan  log  26 

21.6  20.8 

9  33.3 

32.4 

I 

24.3  23.4 

40°  tan  9.9238  9264 

9289 

9315 

9341  9366 

9392  cot  49°  26 

log  cot  0.0762  0736 

0711 

0685 

0659  0634 

0608  tan  log  26 

41°  tan  9.9392  9417 

9443 

9468 

9494  9519 

9544  cot  48°  25 

25 

24 

35 

34 

log  cot  0.0(i08  0583 

0557 

0532 

0506  0481 

0456  tan  log  25 

] 

2  5 

2  4 

1     3.5 

3.4 

42°  tan  9.9544  9570 

9595 

9621 

9646  9671 

9697  cot  47s  25 

50 

"is 

2    7.0 
3  10.5 

6.8 
10.2 

log  cot  0.0456  0430 

0405 

0379 

0354  0329 

0303  tan  log  25 

."5 

7.5 

4  14.0 

13.6 

•1 

10.0 

9.6 

6  17.5 

17.0 

43°  tan  9.9697  9722  9747 

9772 

9798  9823 

9848  cot  46°  25 

5 

12.5 
15  0 

12.0 
14  4 

6  21.0 
7  24  5 

20.4 
23  8 

log  cot  0.0303  0278 

0253 

0228 

0202  0177 

0152  tan  log  25 

7 

\7'.S 

lf.8 

8  28^0 

27^2 

44'  tan  9.9S48  9874 

'.IS'.  Ill 

9924 

9949  9975 

0000  cot  45°  25 

8 

20.0 

19.2 

931.5 

30.6 

log  cot  0.0152  0126 

0101 

0076 

0051  0025  0000  tan  log  25    » 

22.5 

21.6 

45°  tan  0.0000  0025  0051     0076    0101  0126  0152  cot  44°  25 
log  cot  0.0000*9975*9949  *9924  *9899*9874*9848  tan  log  25 


P.P. 


60'       50'     40'       30'        20'      10'       0' 


A"    d. 


P.P. 


Logarithms  of  tangents  and  cotangents,  44°  to  67°. 


506 


UNIFIED  MATHEMATICS 


Log  sin  by  minutes  from  0"  to  9^. 


0 

0  6.4637  7648  9408  *0658  *1627  *2419  *3088  *3668  *4180  *4637  50 
10  7.4637   5051  5429  5777  6099  6398  6678  6942  7190  7425  7648  40 

20   7648 

7859  8061  8255 

8439 

8617 

S7S7 

8951 

*109 

*261 

*408  30 

30  7.9408 

551 

GS!» 

822 

952 

*078 

*200 

*319 

*435 

*548 

*658  20 

40  8.0658 

765 

870 

972 

*072 

*169 

*265 

*358 

*450 

*539 

*627  10 

50  8.1627 

713 

797 

880 

961 

*041 

*119 

*196 

*271 

*346 

*419  0  89 

1 

0  8.2419 

490 

561 

630 

699 

766 

832 

898 

962 

*025 

*088  50 

10  8.3088 

150 

210 

270 

329 

388 

445 

502 

558 

613 

66840 

20   668 

722 

775 

828 

880 

931 

982 

*032 

*082 

*131 

*179  30 

30  8.4179 

227 

275 

322 

368 

414 

459 

504 

549 

593 

637  20 

40   637 

680 

723 

765 

807 

848 

890 

930 

971 

*011 

*050  10 

50  8.5050 

090 

129 

167 

206 

243 

281 

318 

355 

392 

428  088 

2 

0  8.5428 

464 

500 

535 

571 

605 

640 

674 

708 

742 

776  50 

10   776 

809 

842 

875 

907 

939 

.972 

*003 

*035 

*066 

*097  40 

20  8.6097 

128 

159 

189 

220 

250 

279 

309 

339 

368 

397  30 

30   397 

426 

454 

483 

511 

539 

567 

595 

622 

650 

677  20 

40   677 

704 

731 

758 

784 

810 

837 

863 

889 

914 

940  10 

50   940 

965 

991 

*016 

*041 

*066 

*090 

*115 

*140 

*164 

*188  0  87 

3 

0  8.7188 

212 

236 

260 

283 

307 

330 

354 

377 

400 

423  50 

10   423 

445 

468 

491 

513 

535 

557 

580 

602 

623 

645  40 

20   645 

667 

688 

710 

731 

752 

773 

794 

815 

836 

857  30 

30   857 

877 

898 

918 

939 

959 

979 

999 

*019 

*039 

*059  20 

40  8.8059 

078 

098 

117 

137 

156 

175 

194 

213 

232 

251  10 

50   251 

270 

289 

307 

326 

345 

363 

381 

400 

418 

436  086 

4 

0  8.8436 

454 

472 

490 

508 

525 

543 

560 

578 

595 

613  50 

10   613 

'630 

647 

665 

682 

699 

716 

733 

749 

766 

783  40 

20   783 

799 

816 

833 

849 

865 

882 

898 

914 

930 

946  30 

30   946 

962 

978 

994 

*010 

*026 

*042 

*057 

*073 

*089 

*104  20 

40  8.9104 

119 

135 

150 

166 

181 

196 

211 

226 

241 

256  10 

50   256 

271 

286 

301 

315. 

330 

345 

359 

374 

389* 

403  085 

5 

0  8.9403 

417 

432 

446 

460 

475 

489 

503 

517 

531 

545  50 

10   545 

559 

573 

587 

601 

614 

628 

642 

655 

669 

682  40 

20   682 

696 

709 

723 

736 

750 

763 

776 

789 

803 

816  30 

30   816 

829 

842 

855 

868 

881 

894 

907 

919 

932 

945  20 

40   945 

958 

970 

983 

996 

*008 

*021 

*033 

*046 

*058 

*070  10 

50  9.0070 

083 

095 

107 

120 

132 

144 

156 

168 

180 

192  0  84 

6 

0  9.0192 

204 

216 

228 

240 

252 

264 

276 

287 

299 

311  50 

10   311 

323 

334 

346 

357 

369 

380 

392 

403 

415 

426  40 

20   426 

438 

449 

460 

472 

483 

494 

505 

516 

527 

539  30 

30   539 

550 

561 

572 

583 

594 

605 

616 

626 

637 

648  20 

40   648 

659 

670 

680 

691 

702 

712 

723 

734 

744 

755  10 

50   755 

765 

776 

786 

797 

807 

818 

828 

838 

849 

859  0  83 

7 

0  9.0859 

869 

879 

890 

900 

910 

920 

930 

940 

951 

961  50 

10   961 

971 

981 

991 

*001 

*011 

*020 

*030 

*040 

*050 

*06040 

20  9.1060 

070 

080 

089 

099 

109 

118 

128 

138 

147 

157  30 

30   157 

167 

176 

186 

195 

205 

214 

224 

233 

242 

252  20 

40   252 

261 

271 

280 

289 

299 

308 

317 

326 

336 

345  10 

50   345 

354 

363 

372 

381 

390 

399 

409 

418 

427 

436  082 

8 

0  9.1436 

445 

453 

462 

471 

480 

489 

498 

507 

516 

525  50 

10   525 

533 

542 

551 

560 

568 

577 

586 

594 

603 

612  40 

20   612 

620 

629 

637 

646 

655 

663 

672 

680 

689 

697  30 

30   697 

705 

714 

722 

731 

739 

747 

756 

764 

772 

781  20 

40   781 

789 

797 

806 

814 

822 

830 

838 

847 

855 

863  10 

50   863  . 

871 

879 

887 

895 

903 

911 

919 

927 

935 

943  0  81 

10' 


3'         V 


Log  cos  by  minutes  from  81°  to  90°. 


TABLES 


507 


Log  tan  by  minutes  from  (T  to  9°. 


10' 


0 

0       6.4637 
10  7.4637   5051 

7648  9408  *0658 
5429  5777  6099 

*1627 
6398 

*2419 
6678 

*3088 
6942 

*3668 
7190 

*4180 
7425 

*4637  50 
7648  40 

20  7648 

7860 

8062 

8255 

8439 

8617 

8787 

8951 

*109 

*261 

*409  30 

30  7.9409 

00  1 

689 

823 

952 

*078 

*200 

*319 

*435 

*548 

*658  20 

40  8.0658 

765 

870 

972 

*072 

*170 

*265 

*359 

*450 

*540 

*627  10 

50  8.1627 

713 

798 

880 

962 

*041 

*120 

*196 

*272 

*346 

*419  0  89 

1 

0  8.2419 

491 

562 

631 

700 

767 

833 

899 

963 

*026 

*089  50 

in  8.3069 

150 

211 

271 

330 

389 

446 

503 

559 

614 

669  40 

20   669 

723 

776 

829 

881 

932 

983 

*033 

*083 

132 

181  30 

30  8.4181 

229 

276 

323 

370 

416 

461 

506 

551 

595 

638  20 

40   638 

682 

725 

767 

809 

851 

892 

933 

973 

*013 

*053  10 

50  8.5053 

092 

131 

170 

208 

246 

283 

321 

358 

394 

431  088 

2 

0  8.5431 

467 

503 

538 

573 

608 

643 

677 

711 

745 

779  50 

10   779 

812 

845 

878 

911 

943 

975 

*007 

*038 

*070 

*101  40 

20  8.6101 

i:?2 

163 

193 

223 

254 

283 

313 

343 

372 

401  30 

30   401 

430 

459 

487 

515 

544 

571 

599 

627 

654 

682  20 

40   682 

709 

736 

762 

789 

815 

842 

868 

894 

920 

945  10 

50   945 

971 

996 

*021 

*046 

*071 

*096 

*121 

*145 

*170 

*194  0  87 

:{ 

0  8.7194 

218 

242 

266 

290 

313 

337 

360 

383 

406 

429  50 

10    429 

452 

475 

497 

520 

542 

565 

587 

609 

631 

652  40 

20   652 

674 

096 

717 

739 

760 

781 

802 

823 

844 

865  30 

30   865 

886 

906 

927 

947 

967 

988 

*008 

*028 

*048 

*067  20 

40  8.8067 

087 

107 

126 

146 

165 

185 

204 

223 

242 

261  10 

50   261 

280 

299 

317 

336 

355 

373 

392 

410 

428 

446  0  86 

4 

0  8.8446 

465 

483 

501 

518 

536 

554 

572 

589 

607 

624  50 

10   624 

642 

659 

676 

694 

711 

728 

745 

762 

778 

795  40 

20   795 

812 

829 

845 

862 

878 

895 

911 

927 

944 

960  30 

30   960 

976 

992 

*008 

*024 

*040 

*056 

*071 

*087 

*103 

*118  20 

40  8.9118 

134 

150 

165 

180 

196 

211 

226 

241 

256 

272  10 

50   272 

287 

302 

316 

331 

346 

361 

376 

390 

405 

420  0  85 

5 

0  8.9420 

434 

449 

463 

477 

492 

506 

520 

534 

549 

563  50 

10   563 

577 

591 

605 

619 

633 

646 

660 

674 

688 

701  40 

20   701 

7i:> 

729 

742 

756 

769 

782 

796 

809 

823 

836  30 

30   836 

849 

862 

875 

888 

901 

915 

927 

940 

953 

966  20 

40   966 

979 

992 

*005 

*017 

*030 

*O43 

*05o 

*068 

*080 

*093  10 

50  9.0093 

105 

118 

130 

143 

155 

167 

180 

192 

204 

216  084 

(i 

0  9.0216 

228 

240 

253 

265 

277 

289 

300 

312 

324 

336  50 

10   336 

348 

360 

371 

383 

395 

407 

418 

430 

441 

453  40 

20   453 

464 

476 

487 

499 

510 

521 

533 

544 

555 

567  30 

30   567 

578 

688 

600 

611 

622 

633 

645 

656 

667 

678  20 

40   678 

688 

099 

710 

721 

732 

743 

754 

764 

775 

786  10 

50   786 

796 

807 

818 

828 

839 

849 

860 

871 

881 

891  083 

1 

0  9.0891 

902 

912 

923 

933 

943 

954 

964 

974 

984 

995  50 

10   995 

*005 

*015  *025 

*035 

*045 

*055 

*066 

*076 

*086 

*096  40 

20  9.1096 

106 

Lie 

125 

188 

145 

155 

165 

175 

185 

194  30 

:«)   I'M 

204 

214 

223 

233 

243 

252 

262 

272 

281 

291  20 

40   291 

300 

310 

319 

329 

338 

348 

357 

367 

376 

385  10 

50   385 

395 

404 

413 

423 

432 

441 

450 

460 

469 

478  0  82 

s 

0  9.1478 

487 

496 

505 

515 

524 

533 

542 

551 

560 

569  50 

10   569 

687 

696 

605 

613 

622 

631 

640 

649 

658  40 

20   658 

867 

675 

684 

698 

702 

710 

719 

728 

736 

745  30 

30   745 

754 

762 

771 

779 

788 

797 

805 

814 

822 

831  20 

40   831 

839 

848 

856 

864 

873 

881 

890 

898 

906 

915  10 

50   915 

923 

931 

940 

948 

956 

964 

973 

981 

989 

997  0  81 

III 


Log  cot  by  minutes  from  81°  to  90°. 


508 


UNIFIED  MATHEMATICS 


Numerical  values  of  the  sine  function,  0  to  45°. 


10' 


20' 


30' 


40'       60'       60' 


P.P. 


e 

0.0000 

0029 

0058 

0087 

0116 

0145 

0175 

89 

29 

i 

0175 

0204 

0233 

0262 

0291 

0320 

0349 

88 

29 

2 

0349 

0378 

0407 

0436 

0465 

0494 

0523 

87 

29 

30  29 

3 

0523 

0552 

0581 

0610 

0640 

0669 

0698 

86 

29 

1  30  29 

4 

0698 

0727 

0756 

0785 

0814 

0843 

0872 

85 

29 

2  6.0  5.8 

3  9.0  8.7 

5 

0.0872 

0901 

0929 

0958 

0987 

1016 

1045 

84 

29 

4  12.0  11.6 
5  15.0  14.5 

6 

1045 

1074 

1103 

1132 

1161 

1190 

1219 

83 

29 

6  18.0  17.4 

7 

1219 

1248 

1276 

1305 

1334 

1363 

1392 

82 

29 

7  21.0  20.3 

8 

1392 

1421 

1449 

1478 

1507 

1536 

1564 

81 

29 

8  24.0  23.2 

n  97  Q  2fi  1 

9 

1564 

1593 

1622 

1650 

1679 

1708 

1736 

80 

29 

10 

0.1736 

1765 

1794 

1822 

1851 

1880 

1908  N 

79 

29 

28    27 

11 

1908 

1937 

1965 

1994 

2022 

2051 

2079 

78 

28 

12 

2079 

2108 

2136 

2164 

2193 

2221 

2250 

77 

28 

1  2.8  2.7 
2  56  54 

13 

2250 

2278 

2306 

2334 

2363 

2391 

2419 

76 

28 

3  84  8.1 

14 

2419 

2447 

2476 

2504 

2532 

2560 

2588 

75 

28 

4  11.2  10.8 

5  14.0  13.5 

6  16  8  16  2 

15 

0.2588 

2616 

2644 

2672 

2700 

2728 

2756 

74 

28 

7  19.6  18.9 

16 

2756 

2784 

2812 

2840 

2868 

2896 

2924 

73 

28 

8  22.4  21.3 

17 

2924 

2952 

2979 

3007 

3035 

3062 

3090 

72 

28 

9  25.2  24.3 

18 

3090 

3118 

3145 

3173 

3201 

3228 

3256 

71 

28 

19 

3256 

3283 

3311 

3338 

3365 

3393 

3420 

70 

27 

26   25 

20 

0.3420 

3448 

3475 

3502 

3529 

3557 

3584 

69 

27 

1  2.6  2.5 

21 

3584 

3611 

3638 

3665 

3692 

3719 

3746 

68 

27 

2  5.2  5.0 

22 

3746 

3773 

3800 

3827 

3854 

3881 

3907 

67 

27 

3  7.8  7.5 
4  10  4  10  0 

23 

3907 

3934 

3961 

3987 

4014 

4041 

4067 

66 

27 

5  13  0  12.5 

24 

4067 

4094 

4120 

4147 

4173 

4200 

4226 

65 

26 

6  15.6  15.0 

7  18.2  17.5 

30' 

8  20.8  20.0 

9  23.4  22.5 

25 

0.4226 

4253 

4279 

4305 

4331 

4358 

4384 

64 

26 

26 

4384 

4410 

4436 

4462 

4488 

4514 

4540 

63 

26 

27 

4540 

4566 

4592 

4617 

4643 

4669 

4695 

62 

26 

24  23 

28 

4695 

4720 

4746 

4772 

4797 

4823 

4848 

61 

26 

1  2.4  2.3 

29 

4848 

4874 

4899 

4924 

4950 

4975 

5000 

60 

25 

2  4.8  4.6 

3  7.2  6.9 

4  96  9.2 

30 

0.5000 

5025 

5050 

5075 

5100 

5125 

5150 

59 

25 

5  12.0  11.5 

31 

5150 

5175 

5200 

5225 

5250 

5275 

5299 

58 

25 

6  14.4  13.8 

32 

5299 

5324 

5348 

5373 

5398 

5422 

5446 

57 

24 

7  16.8  16.1 

33 

5446 

5471 

5495 

5519 

5544 

5568 

5592 

56 

24 

8  19.2  18.4 
9  21.6  20.7 

34 

5592 

5616 

5640 

5664 

5688 

5712 

5736 

55 

24 

35 

0.5736 

5760 

5783 

5807 

5831 

5854 

5878 

54 

24 

22   21  20 

36 

5878 

5901 

5925 

5948 

5972 

5995 

6018 

53 

23 

37 

6018 

6041 

6065 

6088 

6111 

6134 

6157 

52 

23 

1  2.2  2.1  2.0 
2  44  42  40 

38 

6157 

6180 

6202 

6225 

6248 

6271 

6293 

51 

23 

3  6^6  6.3  6.0 

39 

6293 

6316 

6338 

6361 

6383 

6406 

6428 

50 

22 

4  8.8  8.4  8.0 

5  11.0  10.5  10.0 

6  13.2  12.6  12.0 

40 

0.6428 

6450 

6472 

6494 

6517 

6539 

6561 

49 

22 

7  15.4  14.7  14.0 

41 

6561 

6583 

6604 

6626 

6648 

6670 

6691 

48 

22 

8  17.6  16.8  16.0 

42 

6691 

6713 

6734 

6756- 

6777 

6799 

6820 

47 

22 

9  19.8  18.9  18.0 

43 

6820 

6841 

6862 

6884 

6905 

6926 

6947 

46 

21 

44 

6947 

6967 

6988 

7009 

7030 

7050 

7071 

45 

21 

60' 


40' 


30' 


20'       10' 


d. 


P.P. 


Numerical  values  of  the  cosine  function,  45°  to  90°. 


TABLES 


509 


Numerical  values  of  the  sine  function,  45°  to  90°. 


(K       10'      20' 


30' 


40'        50'        60' 


(I 


P.  P. 


45 

0.7071 

7092 

7112 

7133 

7153 

7173 

7193 

44 

20 

M 

7193 

7214 

7234 

7254 

7274 

7294 

7314 

43 

20 

21  20  1»  18 

47 

7314 

7:«3 

7353 

7373 

7392 

7412 

7431 

4',' 

20 

1 

21  20  19  18 

4S 

7431 

7451 

7470 

7490 

7509 

7528 

7547 

41 

19 

•2 

4.2  4.0  3.8  3.6 

49 

7547 

7566 

7585 

7604 

7623 

7642 

7660 

40 

19 

1 

6.3  6.0  5.7  5.4 

1 

8.4  8.0  7.6  7.2 

5 

10.5  10.0  9.5  9.0 

51) 

0.7660 

7679 

7698 

7716 

7735 

7753 

7771 

3» 

18 

a 

12.6  12.0  11.4  10.8 

51 

n 

5:1 

7771 
7880 
7986 

7790 
7898 
8004 

7808 
7916 
8021 

7826 
7934 
8039 

7844 
7951 
8056 

7862 
7969 
8073 

7880 
7986 
8090 

:is 
37 

::<; 

18 
18 
17 

7 

s 

!» 

14  7  14.0  13.3  12.6 
16.8  16.0  15.2  14.4 
18.9  18.0  17.1  16.2 

54 

8090 

8107 

8124 

8141 

8158 

8175 

8192 

35 

17 

17   16  15  14 

H 

0.8192 

8208 

8225 

8241 

8258 

8274 

8290 

34 

16 

1 

1.7  1.6  1.5  1.4 

5*i 

8290 

S307 

8323 

8339 

8355 

8371 

8387 

:w 

16 

1 

3.4  3.2  3.0  2.8 

57 

8387 

8403 

8418 

8434 

8450 

8465 

'8480 

3'i 

16 

:i 
4 

5.1  4.7  4.5  4.2 
68  64  60  56 

M 

8480 

8496 

8511 

8526 

8542 

8557 

8572 

31 

15 

5 

8.5  S.O  7.5  7.0 

m 

8572 

8587 

8601 

8616 

B831 

8646 

8660 

30 

15 

i 

10.2  9.6  90  8.4 

7 

11.9  11.2  10.5  9.8 

8 

13.6  12.8  12.0  11.2 

60 

0.8660 

8675 

8689 

8704 

8718 

8732 

8746 

','9 

14 

i 

15.3  14.4  13.5  12.6 

61 

8746 

8760 

8774 

8783 

8802 

8816 

8829 

•>s 

14 

n 

8829 

ssi:i 

8857 

8870 

8884 

8897 

8910 

•iJ 

14 

<i3 

8910 

8923 

8936 

8949 

8962 

8975 

8988 

•><; 

13 

13  12  11  10 

H 

8988 

9001 

9013 

9026 

9038 

9051 

9063 

25 

12 

i 

1.3  1.2  1.1  1.0 

i 

2.6  2.4  2.2  2.0 

:i 

3.9  3.6  3.3  3.0 

H 

0.9063 

9075 

9088 

9100 

9112 

9124 

9135 

•>4 

12 

-I 

5.2  4.8  4.4  4.0 

N 

9135 

9147 

9159 

9171 

9182 

9194 

9205 

23 

12 

5 

6.5  6.0  55  5.0 

«7 
6H 

020:> 
9272 

9216 
9283 

9228 
9293 

9239 
9304 

9250 
9315 

9261 

O.S2.-> 

9272 
9336 

n 
21 

11 
11 

a 

7 
• 

7.8  7.2  6.6  6.0 
9.1  8.4  7.7  7.0 
104  96  88  80 

6» 

9336 

9346 

9356 

9367 

9377 

9387 

9397 

•io 

10 

1 

11.7  10.8  9.9  9.0 

307 

70 

0.9397 

9407 

9117 

9426 

9436 

9446 

9455 

19 

10 

71 

9455 

'.  14  (•,:> 

9474 

9483 

9492 

9502 

9511 

IS 

9 

n 

9511 

9520 

9528 

9537 

9546 

9555 

9563 

17 

9 

73 

9563 

9572 

9.-.SO 

9588 

9596 

9605 

9613 

16 

8 

74 

9613 

9621 

9628 

9636 

9644 

9652 

9659 

15 

8 

75 

0.9659 

9667 

9674 

9681 

9689 

9696 

9703 

14 

7 

7<; 

9703 

9710 

9717 

9724 

!I7:50 

9737 

9744 

13 

7 

77 

9744 

9750 

9757 

9763 

9769 

9775 

9781 

n 

6 

7S 

9781 

9787 

9793 

9799 

9805 

9811 

9816 

n 

6 

n 

9816 

9822 

9827 

9833 

9838 

9843 

9848 

10 

5 

Interpolate  men- 

tally,  using  the  mul- 

89 

0.9848 

9853 

9858 

9863 

9868 

9872 

9877 

9 

5 

tiplication  table. 

SI 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

8 

4 

S'» 

9903 

Wl>7 

9911 

9914 

9918 

9922 

9925 

7 

4 

S3 

rasa 

<t'.)2<> 

9932 

9936 

9939 

9942 

994f> 

6 

3 

S4 

9945 

9948 

9951 

9954 

9957 

9959 

9962 

& 

3 

S5 

0.9962 

9964 

9967 

9969 

•  9971 

9974 

9976 

4 

2 

H6 

9976 

9978 

9980 

9981 

9983 

9985 

9986 

1 

2 

S7 

M86 

9988 

9989 

9990 

9992 

9993 

9994 

1 

1 

HS 

'.!•''!.-, 

9996 

9997 

9997 

9998 

9998 

1 

1 

S9 

MNM 

9999 

9999 

*0000 

*0000 

*0000 

*0000 

0 

0 

60'      50'      40' 


30' 


W 


(1 


P.P. 


Numerical  values  of  the  cosine  function,  0  to  45°. 


510 


UNIFIED  MATHEMATICS 


Numerical  values  of  the  tangent  function,  0°  to  45°. 


0'        10'       20'        30'        40'       50'       60' 


d. 


P.  P. 


0 

0.0000 

0029 

0058 

0087 

0116 

0145 

0175 

89 

29 

29  30  31  32  33 

1 

0175 

0204 

0233 

0262 

0291 

0320 

0349 

88 

29 

1  2.9  3.0  3.1  3.2  3.3 

2 

0349 

0378 

0407 

0437 

0466 

0495 

0524 

87 

29 

2  5.8  6.0  6.2  6.4  6.6 

3 

0524 

0553 

0582 

0612 

0641 

0670 

0699 

86 

29 

3  8.7  9.0  9.3  9.6  9.9 

4 

0699 

0729 

0758 

0787 

0816 

0846 

0875 

85 

29 

4  11.6  12.0  12.4  12.8  13.2 
5  14.5  15.0  15.5  16.0  16.5 

6  17.4  18.0  18.6  19.2  19.8 

5 

6 

0.0875 
1051 

0904 
1080 

0934 
1110 

0963 
1139 

0992 
1169 

1022 
1198 

1051 

1228 

84 
83 

29 
30 

7  20.3  21.0  21.7  22.4  23.1 
8  23.2  24.0  24.8  25.6  26.4 
9  26.1  27.0  27.9  28.8  29.7 

7 

1228 

1257 

1287 

1317 

1346 

1376 

1405 

82 

30 

8 

1405 

1435 

1465 

1495 

1524 

1554 

1584 

81 

30 

34  35  36  37  38 

9 

1584 

1614 

1644 

1673 

1703 

1733 

1763 

80 

30 

1  3.4  3.5  3.6  3.7  3.8 

2  6.8  7.0  7.2  7.4  7.6 

10 
11 

0.1763 
1944 

1793 

1974 

1823 
2004 

1853 
2035 

1883 
2065 

1914 
2095 

1944 
2126 

79 

78 

30 
30 

3  10.2  10.5  10.8  11.1  11.4 
4  13.6  14.0  14.4  14.8  15.2 
5  17  0  17  5  18  0  18  5  19  0 

12 

2126 

2156 

2186 

2217 

2247 

2278 

2309 

77 

30 

6  20.4  21.0  21.6  22.2  22.8 

13 

2309 

2339 

2370 

2401 

2432 

2462 

2493 

76 

31 

7  23.8  24.5  25.2  25.0  26.6 

14 

2493 

2524 

2555 

2586 

2617 

2648 

2679 

75 

31 

8  27.2  28.0  28.8  29.6  30.4 

9  30.6  31.5  32.4  33.3  34.2 

15 

0.2679 

2711 

2742 

2773 

2805 

2836 

2867 

74 

31 

39  40  41  42  43 

16 
17 

2867 
3057 

2899 
3089 

2931 
3121 

3962 
3153 

2994 
3185 

3026 
3217 

3057 
3249 

73 
72 

32 
32 

1  3.9  4.0  4.1  4.2  4.3 

2  78  80  82  84  86 

18 

3249 

3281 

3314 

3346 

3378 

3411 

3443 

71 

32 

3  11.7  12'.0  12.3  12.6  12.9 

19 

3443 

3476 

3508 

3541 

3574 

3607 

3640 

70 

33 

4  15.6  16.0  16.4  16.8  17.2 

5  19.5  20.0  20.5  21.0  21.5 

6  23.4  24.0  24.6  25.2  25.8 

20 

0.3640 

3673 

3706 

3739 

3772 

3805 

3839 

69 

33 

7  27.3  28.0  28.7  29.4  30.1 

21 

3839 

3872 

3906 

3939 

3973 

4006 

4040 

68 

34 

8  31.2  32.0  32.8  33.6  34.4 

22 

4040 

4074 

4108 

4142 

4176 

4210 

4245 

67 

34 

9  35.1  36.0  36.9  37.8  38.7 

23 

4245 

4279 

4314 

4348 

4383 

4417 

4452 

66 

34 

24 

4452 

4487 

4522 

4557 

4592 

4628 

4663 

65 

35 

44  45  46  47  48 

1  4.4  4.5  4.6  4.7  4.8 

30' 

2  8.8  9.0  9.2  9.4  9.6 

3  13.2  13.5  13.8  14.1  14.4 

25 

0.4663 

4699 

4734 

4770 

4806 

4841 

4877 

64 

36 

4  17.6  18.0  18.4  18.8  19.2 
5  22  0  22  5  23  0  23  5  24.0 

26 

4877 

4913 

4950 

4986 

5022 

5059 

5095 

63 

36 

6  26.4  27.0  27.6  28.2  28.8 

27 

5095 

5132 

5169 

5206 

5243 

5280 

5317 

62 

37 

7  30.8  31.5  32.2  32.9  33.6 

28 
29 

5317 
5543 

5354 
5581 

5392 
5619 

5430 
5658 

5467 
5696 

5505 
5735 

5543 
5774 

61 
60 

38 
38 

8  35.2  36.0  36.8  37.6  38.4 
9  39.6  40.5  41.4  42.3  43.2 

49  50  51  52  53 

30 

0.5774 

5812 

5851 

5890 

5930 

5969 

6009 

59 

39 

1  4.9  5.0  5.1  5.2  5.3 

31 

6009 

6048 

6088 

6128 

6168 

6208 

6249 

58 

40 

2  9^8  lo!o  10'2  10^4  lo!6 

32 

6249 

6289 

6330 

6371 

6412 

6453 

6494 

57 

41 

3  14.7  15.0  15.3  15.6  15.9 

33 

6494 

6536 

6577 

6619 

6661 

6703 

6745 

56 

42 

4  19.6  20.0  20.4  20.8  21.2 

34 

6745 

6787 

6830 

6873 

6916 

6959 

7002 

55 

43 

5  24.5  25.0  25.5  26.0  26.5 
6  29.4  30.0  30.6  31.2  31.8 

7  34.3  35.0  35:7  36.4  37.1 

35 
36 

0.7002 
7265 

7046 
7310 

7089 
7355 

7133 
7400 

7177 
7445 

7221 
7490 

7265 
7536 

54 
53 

44 
45 

8  39.2  40.0  40.8  41.6  42.4 
9  44.1  45.0  45.9  46.8  47.7 

37 

7536 

7581 

7627 

7673 

7720 

7766 

7813 

52 

46 

38 

7813 

7860 

7907 

7954 

8002 

8050 

8098 

51 

48 

54  55  56  57  58 

39 

8098 

8146 

8195 

8243 

8292 

8342 

8391 

50 

49 

1  5.4  5.5  5.6  5.7  5.8 

2  10.8  11.0  11.2  11.4  11.6 

3  16.2  16.5  16.8  17.1  17.4 

40 

0.8391 

8441 

8491 

8541 

8591 

8642 

8693 

49 

50 

4  21.6  22.0  22.4  22.8  23.2 

41 
42 
43 
44 

8693 
9004 
9325 
9657 

8744 
9057 
9380 
9713 

8796 
9110 
9435 
9770 

8847 
9163 
9490 
9827 

8899 
9217 
9545 
9884 

8952 
9271 
9601 
9942 

9004 
9325 
9657 
*0000 

48 
47 
46 
45 

52 
54 
55 
57 

5  27.0  27.5  28.0  28.5  29.0 
6  32.4  33.0  33.6  34.2  34.8 
7  37.8  38.5  39.2  39.9  40.6 
8.43.2  44.0  44.8  45.6  46.4 
9  48.6  49.5  50.4  51.3  52.2 

60' 

50' 

40' 

30' 

20' 

IV 

0' 

° 

d. 

P.P. 

Numerical  values  of  the  cotangent  function,  45°  to  90°. 


TABLES 


511 


Numerical  values  of  the  tangent  function,  45°  to  90°. 


10' 


20' 


30' 


40' 


50' 


60' 


d. 


P.P. 


45    1.000    1.006    1.012 

1.018 

1.024 

1.030    1.03644    6 

6     7     8     9    10  11 

46    1.036    1.042    l.OIS 

1.054 

1.060 

1.066    1.07243    6 

1060708091011 

47    1.072    1.079    1.085 

1.091 

1.098 

1.104    1.11142    6 

2  1.2  1.4  1.8  1.8  2.0  2.2 

48    1.111    1.117    1.124 

1.130 

1.137 

1.144    1.15041    6 

3  1.8  2.1  2.4  2.7  3.0  3.3 

4»    1.150    1.157    1.104 

1.171 

1.178 

1.185    1.19240    7 

4  2.4  2.8  3.2  3.6  4.0  4.4 

5  3.0  3.5  4.0  4.5  5.0  5.5 

6  3.6  4.2  4.8  5.4  6.0  6.6 

50    1.192    1.199    1.206 
51    1.235    1.242    1.250 
52    1.280    1.288    1.295 

1.213 
1.257 
1.303 

1.220 
1.265 
1.311 

1.228    1.23539    7 
1.272    1.28038    8 
1.319    1.32737    8 

7  4.2  4.9  5.6  6.3  7.0  7.7 
8  4.8  5.6  6.4  7.2  8.0  8.8 
9  5.4  6.3  7.2  8.1  9.0  9.9 

53    1.327    1.335    1.343 

1.351 

1.360 

1.368    1.37636    8 

54    1.376    1.385    1.393 

1.402 

1.411 

1.419    1.42835    9 

12    13    14    15    16 

1    1.2    1.3    1.4    1.5    1.6 

2    2.4    2.6    2.8    3.0    3.2 

55    1.428    1.437    1.446 

1.455 

1.464 

1.473    1.48334    9 

3    3.6    3.9    4.2    4.5    4.8 

56     1.IS3     1.492     1.501 

i.an 

1.520 

1.530    1.5403310 

4    4.8    5.2    5.6    6.0    6.4 

57    1.540    1.550    1.560 

1.570 

1.580 

1.500    1.6003210 

5    6.0    6.5    7.0    7.5    8.0 
6    72    78    84    90    96 

58    1.600    1.611    1.021 

1.632 

1.643 

1.653    1.66431  11 

7    8^4    9.1    9.8  10.5  11.2 

59    1.664    1.675    1.686 

1.698 

1.709 

1.720    1.7323011 

8    9.6  10.4  11.2  12.0  12.8 

9  10.8  11.7  12.6  13.5  14.4 

60    1.732    1.744    1.756 

1.767 

1.780 

1.792    1.804  29  12 

n|w        1O        *>ll        91 

61    1.804    1.816    1.829 

1.842 

1.855 

1.868    1.881  28  13 

19        1  •'        £V         .  1 

62    1.881    1.894    1.907 

1.921 

1.935 

1.949    1.963  27  14 

1    1.7    1.8    1.9    2.0    2.1 

63    1.963    1.977    1.991 

2.006 

2.020 

2.035    2.050  26  14 

2    3.4    3.6    3.8    4.0    4.2 
3    51     54    57    60    63 

6ft    2.050    2.066    2.081 

2.097 

2.112 

2.128    2.145  25  16 

4    6^8    7^2    7^6    8.0    8.4 

5    8.5    9.0    9.5  10.0  10.5 

6102108114120  12  6 

65    2.145    2.161    2.177 

2.194 

2.211 

2.229    2.246  24  17 

7  11.9  12.6  13.3  14.014.7 

66    2.246    2.264    2.282 

2.300 

2.318 

2.337    2.356  23  18 

8  13.6  14.4  15.2  16.0  16.8 

67    2.356    2.375    2.394 

2.414 

2.434 

2.455    2.475  22  20 

9  15.3  16.2  17.1  18.0  18.9 

68    2.475    2.496    2.517 

2.539 

2.560 

2.583    2.6052122 

69    2.605    2.628    2.651 

2.675 

2.699 

2.723    2.7472024 

22    23    2*    25    26 

1    2.2    2.3    2.4    2.5    26 

30' 

2    4.4    4.6    4.8    5.0    5.2 

3    6.6    6.9    7.2    7.5    7.8 

76   2.747    2.773    2.798 
71    2.904    2.932    2.000 

2.824 
2.989 

2.850 
3.018 

2.877    2.9041926 
3.047    3.078  18  29 

4    8.8    9.2    9.6  10.0  10.4 
5  11.011.5  12.012.5  13.0 
6  13.2  13.8  14.4  15.0  15.6 

72    3.078    3.108    3.140 

3.172 

3.204 

3.237    3.271  17  32 

7  15.4  16.1  16.8  17.5  18.2 

73    3.271    3.305    3.340 

3.376 

3.412 

3.450    3.487  16  36 

8  17.6  18.4  19.2  20.020.8 

74    3.487    3.526    3.566 

3.606 

3.647 

3.689    3.7321541 

9  19.820.721.622.523.4 

27    28    59    62    63 

75   3.732    3.776    3.821 

3.867 

3.914 

3.962    4.0111446 

76    4.011    4.061    4.113 
77    4.331    4.390    4.449 

4.165 
4.511 

4.219 
4.574 

4.275    4.331  13  53 
4.638    4.7051262 

1     2.7    2.8    5.9    6.2    6.3 
2    5.4    5.6  11.8  12.4  12.6 
3    8.1    8.4  17.7  18.6  18  9 

78    4.705    4.773    4.843 

4.915 

4.989 

5.066    5.1451173 

4  10.8  11.2  23.624.825.2 

7t    5.145    5.226    5.309 

5.396 

5.485 

5.576    5.671  10  88 

5  13.5  14.029.531.031.5 

6  16.2  16.8  35.4  37.2  37.8 

7  18.9  19.641.3  43.444.1 

80   5.671    5.769    5.871 

5.976 

6.084 

6.197    6.314    9 

8  21.  6  22.4  47.2  49.650.4 

81    6.314    6.435    6.561 

6.691 

6.827 

6.968    7.115    8 

9  24.3  25.2  53.1  55.8  56.7 

H'J    7.115    7.269    7.429 

7.596 

7.770 

7.953    8.144    7 

83    8.144    8.345    8.556 

8.777 

9.010 

9.255    9.514    6 

64     66     68    70     72 

84    9.514    9.78810.078 

10.385 

10.712 

11.059  11.430    5 

1    6.4    6.6    6.8    7.0    7.2 

2  12.8  13.2  13.6  14.0  14.4 

f 

3  19.2  19.8  20.421.021.6 

8511.43011.82612.251 

12.706 

13.197 

13.72714.301    4 

425.626.427.228.028.8 

86  14.301  14.1)24  15.<;05 
,H7  1  d.  nsi  20.206  21.470 
HM2s.o:«i  31.242  34.368 

16.350 
22.904 

iiS.ISS 

17.169 
24.542 
42.964 

18.075  10.  OS1     :i 
26.43228.636    2 
49.10457.290    1 

5  32.0  33.0  34.0  35.0  36.0 
6  38.4  39.6  40.8  42.0  43.2 
7  44.8  46.2  47.6  49.0  50.4 
8  51  .2  52.8  54.4  56  0  57  6 

8957.29068.75085.940 

114.59 

171.89 

343.77    infinit.  0 

957.659.461.263.064.8 

<>o         50'       40' 


30' 


20' 


10' 


P.P. 


Numerical  values  of  the  cotangent  function,  0°  to  45°. 


512 


rNIFIKI)    MATHEMATICS 


Radian  measure  of  angles,  0D  to  ISO"1 

or 
Length  of  arc  in  unit  circle  for  angle  Q-  to  180D  at  center. 


A"       RADIANS 


A°       RADIANS 


RADIANS 


A°        RADIANS 


1° 

2° 
3° 

0.017 
0.035 
0.052 

4C° 

47° 
48° 

0.803 
0.820 
0.838 

91° 

92° 
93° 

1.588 
1.606 
1.623 

136° 
137° 
138° 

2.374 
2.391 
2.409 

4° 
5° 
6° 

0.070 
0.087 
0.105 

49° 
50° 
51° 

0.855 
0.873 
0.890 

94° 
95° 
96° 

1.641 

1.658 
1.676 

139° 
140° 
141° 

2.426 
2.443 
2.461 

7° 
8° 
9° 

0.122 
0.140 
0.157 

52° 
53° 

54° 

0.908 
0.925 
0.942 

97° 
98° 
99° 

1.693 
1.710 
1.728 

142° 

143° 
144° 

2.478 
2.496 
2.513 

10° 
11° 
12° 

0.175 
0.192 
0.203 

55° 
56° 

57° 

0.960 
0.977 
0.995 

100° 
101° 
102° 

1.745 
1.763 
1.780 

145° 
146° 
147° 

2.531 
2.548 
2.566 

13° 
14° 
15° 

0.227 
0.244 
0.262 

58° 
59° 
60° 

1.012 
1.031 
1.047 

103° 
104° 
105° 

1.798 
1.815 
1.833 

148° 
149° 
150° 

2.583 
2.601 
2.618 

16° 
17° 

18° 

0.279 
0.297 
0.314 

61° 
62° 
63° 

1.065 
1.082 
1.100 

106° 
107° 
108° 

1.850 
1.868 
1.885 

151° 
152° 
153° 

2.635 
2.653 
2.670 

19° 
20° 
21° 

0.332 
0.349 
0.367 

64° 
65° 
66° 

1.117 
1.134 
1.152 

109° 
110° 
111° 

1.902 
1.920 
1.937 

154° 

1.55° 
156° 

2.688 
2.705 
2.723 

22° 
23° 
24° 

0.384 
0.401 
0.419 

67° 
68° 
69° 

1.169 
1.187 
1.204 

112° 
113° 
114° 

1.955 
1.972 
1.990 

157° 
158° 
159° 

2.740 
2.758 

2.775 

25° 
26° 
27° 

0.436 
0.454 
0.471 

70° 
71° 

72° 

1.222 
1.239 
1.257 

115° 

116° 
117° 

2.007 
2.025 
2.042 

160° 
161° 
162° 

2.793 
2.810 
2.827 

28° 
29° 
30° 

0.189 
0.506 
0.524 

73° 
74° 
75° 

1.274 
1.292 
1.309 

118° 
119° 
120° 

2.059 
2.077 
2.094 

163° 

164° 
165° 

2.845 
2.862 
2.880 

31° 
32° 
33° 

0.541 
0.559 
0.576 

76° 
77° 
78° 

1.326 
1.344 
1.361 

121° 
122° 
123° 

2.112 
2.129 
2.147 

166° 
167° 
168° 

2.897 
2.915 
2.932 

34° 
35° 
36° 

0.593 
0.611 
0.628 

79° 
80° 
81° 

1.379 
1.396 
1.414 

124° 

125° 
126° 

2.164 
2.182 
2.199 

169° 
170° 
171° 

2.950 
2.967 
2.985 

37° 
38° 
39° 

0.646 
0.663 
0.681 

82° 
83° 
84° 

1.431 
1.449 
1.466 

127° 
128° 
129° 

2.217 
2.234 
2.251 

172° 
173° 
174° 

3.002 
3.019 
3.037 

40° 
41° 
42° 

0.698 
0.716 
0.733 

85° 
86° 
87° 

1.484 
1.501 
1.518 

130° 
131° 
132° 

2.269 
2.286 
2.304 

175° 
176° 
177° 

3.054 
3.072 
3.089 

43° 
44° 
45° 

0.750 
0.768 
0.785 

88° 
89° 
90° 

1.536 
1.553 
1.571 

133° 

134° 
135° 

2.321 
2.339 
2.356 

178° 
179° 
180° 

3.107 
3.124 
3.142 

TABLES 


513 


Minutes  as 

Decimals  of 

Growth  Function,  e* 

One  Degree  or 

Seconds 

Decay  Function,  e~* 

as 

Decimals 

of    One 

e* 

OR  tf 

«~~*  OB 

e-« 

Minute 

X 

t 

log.  i 

logc  t 

Value 

loglo 

Value 

log* 

1 

.017 

31 

.517 

0.0 



oo          1.000 

0.000 

1.000 

0.000 

2 

.033 

32 

.533 

0.1 

-2.303 

1.105 

0.043 

0.905 

9.957 

3 

.050 

33 

.550 

0.2 

-1.610 

1.221 

0.087 

0.819 

9.913 

4 

.067 

34 

.567 

0.3 

-1.204 

1.350 

0.130 

0.741 

9.870 

5 

.083 

35 

.583 

0.4 

-0.916 

1.492 

0.174 

0.670 

9.826 

6 

.100 

36 

.600 

0.5 

-  0.693 

1.649 

0.217 

0.607 

9.783 

7 

.117 

37 

.617 

0.6 

-0.511 

1.822 

0.261 

0.549 

9.739 

8 

.133 

38 

.633 

0.7 

-0.357 

2.014 

0.304 

0.497 

9.696 

9 

.150 

39 

.650 

0.8 

-0.223 

2.226 

0.347 

0.449 

9.653 

10 

.167 

40 

.667 

0.9 

-0.105 

2.460 

0.391 

0.407 

9.609 

11 

.183 

41 

.683 

1.0 

0.000 

2.718 

0.434 

0.368 

9.566 

12 

.200 

42 

.700 

1.1 

0.095 

3.004 

0.478 

0.333 

9.522 

13 

.217 

43 

.717 

1.2 

0.182 

3.320 

0.521 

0.301 

9.479 

14 

.233 

44 

.733 

1.3 

0.262 

3.769 

0.565 

0.273 

9.435 

15 

.250 

45 

.750 

1.4 

0.336 

4.055 

0.608 

0.247 

9.392 

16 

.267 

46 

.767 

1.5 

0.405 

4.482 

0.651 

0.223 

9.349 

17 

.283 

47 

.783 

1.6 

0.470 

4.953 

0.695 

0.202 

9.305 

18 

.300 

48 

.800 

1.7 

0.531 

5.474 

0.738 

0.183 

9.262 

19 

.317 

49 

.817 

1.8 

0.588 

6.050 

0.782 

Q.165 

9.218 

20 

.333 

50 

.833 

1.9 

0.642 

6.686 

0.825 

0.150 

9.175 

21 

.350 

51 

.850 

2.0 

0.693 

7.389 

0.869 

0.135 

9.131 

22 

.367 

52 

.867 

2.1 

0.742 

8.166 

0.912 

0.122 

9.088 

23 

.383 

53 

.883 

2.2 

0.788 

9.025 

0.955 

0.111 

9.045 

24 

.400 

54 

.900 

2.3 

0.833 

9.974 

0.999 

0.100 

0.001 

25 

.417 

55 

.917 

2.4 

0.875 

11.02 

1.023 

0.091 

8.958 

26 

.433 

56 

.933 

2.5 

0.916 

12.18 

1.086 

0.082 

8.914 

27 

.450 

57 

.950 

2.6 

0.956 

13.46 

1.129 

0.074 

8.871 

28 

.467 

58 

.967 

2.7 

0.993 

14.88 

1.173 

0.067 

8.827 

29 

.483 

59 

.983 

2.8 

1.030 

16.44 

1.216 

0.061 

8.784 

30 

.500 

60 

1.000 

2.9 

1.065 

18.17 

1.259 

0.055 

8.741 

3.0 

1.099 

20.09 

1.303 

0.050 

8.697 

3.1 

1.132 

22.20 

1.346 

0.045 

8.654 

3.2 

1.163 

24.53 

1.390 

0.041 

8.610 

3.3 

1.193 

27.11 

1.433 

0.037 

8.567 

3.4 

1.224 

29.96 

1.477 

0.033 

8.523 

3.5 

1.253 

33.12 

1.520 

0.030 

8.480 

4.0 

1.386 

54.60 

1.737 

0.018 

8.263 

4.5 

1.504 

90.02 

1.954 

0.0111 

8.046 

5.0 

1.609 

148.4 

2.171 

0.0067 

7.829 

6.0 

1.792 

403.4 

2.606 

0.0025 

7.394 

7.0 

1.946 

1096.6 

3.040 

0.0009 

6.960 

8.0 

2.079 

2981.0 

3.474 

0.0003 

6.526 

9.0 

2.197 

8103.1 

3.909 

0.0001 

6.091 

10.0 

2.303 

22026. 

4.343 

0,0000 

5.657 

514  UXIFIED  MATHEMATICS 

The  accumulation  of  1  at  the  end  of  n  years.    rn  =  (1  +  i)M. 


Years. 

li%. 

*%. 

**%. 

3%. 

4%. 

5%. 

6%. 

Years. 

1 

1.0150 

1.0200 

1.0250 

1.0300 

1.0400 

1.0500 

1.0600 

1 

2 

1.0302 

1.0404 

1.0506 

1.0609 

1.0816 

1.1025 

1.1236 

2 

3 

1.0457 

1.0612 

1.0769 

1.0927 

1.1249 

1.1576 

1.1910 

1 

4 

1.0614 

1.0824 

1.1038 

1.1255 

1.1000 

1.2155 

1.2625 

4 

5 

1.0773 

1.1041 

1.1314 

1.1593 

1.2167 

1.2763 

1.3382 

5 

6 

1.0934 

1.1262 

1.1597 

1.1941 

1.2653 

1.3401 

1.4185 

6 

7 

1.1098 

1.1487 

1.1887 

1.2299 

1.3159 

1.4071 

1.5036 

7 

8 

1.1265 

1.1717 

1.2184 

1.2668 

1.3686 

1.4775 

1.5938 

8 

9 

1.1434 

1.1951 

1.2489 

1.3048 

1.4233 

1.5513 

1.6895 

9 

10 

1.1605 

1.2190 

1.2801 

1.3439 

1.4802 

1.6289 

1.7908 

10 

11 

1.1779 

1.2434 

1.3121 

1.3842 

1.5395 

1.7103 

1.8983 

11 

12 

1.1956 

1.2682 

1.3449 

1.4258 

1.6010 

1.7959 

2.0122 

12 

13 

1.2136 

1.2936 

1.3785 

1.4685 

1.6651 

1.8856 

2.1329 

13 

14 

1.2318 

1.3195 

1.4130 

1.5126 

1.7317 

1.9799 

2.2609 

14 

15 

1.2502 

1.3459 

1.4483 

1.5580 

1.8009 

2.0789 

2.3966 

15 

16 

1.2690 

1.3728 

1.4845 

1.6047 

1.8730 

2.1829 

2.5404 

16 

17 

1.2880 

1.4002 

1.5216 

1.6528 

1.9473 

2.2920 

2.6928 

17 

18 

1.3073 

1.4282 

1.5597 

1.7024 

2.0258 

/  2.4066 

2.8543 

18 

19 

1.3270 

1.4568 

1.5987 

1.7535 

2.1068 

2.5270 

3.0256 

19 

20 

1.3469 

1.4859 

1.6386 

1.8061 

2.1911 

2.6533 

3.2071 

20 

21 

1.3671 

1.5157 

1.6796 

1.8603 

2.2788 

2.7860 

3.3996 

21 

22 

1.3876 

1.5460 

1.7216 

1.9161 

2.3699 

2.9253 

3.6035 

22 

23 

1.4084 

1.5769 

1.7646 

1.9736 

2.4647 

3.0715 

3.8197 

23 

24 

1.4295 

1.6084 

1.8087 

2.0328 

2.5633 

3.2251 

4.0489 

24 

25 

1.4509 

1.6406 

1.8539 

2.0938 

2.6658 

3.3864 

4.2919 

25 

26 

1.4727 

1.6734 

1.9003 

2.1566 

2.7725 

3.5557 

4.5494 

26 

27 

1.4948 

1.7069 

1.9478 

2.2213 

2.8834 

3.7335 

4.8223 

27 

28 

1.5172 

1.7410 

1.9965 

2.2879 

2.9987 

3.9201 

5.1117 

28 

29 

1.5400 

1.7758 

2.0464 

2.3566 

3.1187 

4.1161 

5.4184 

29 

30 

1.5631 

1.8114 

2.0976 

2.4273 

3.2434 

4.3219 

5.7435 

30 

31 

1.5865 

1.8476 

2.1500 

2.5001 

3.3731 

4.5380 

6.0881 

31 

32 

1.6103 

1.8845 

2.2038 

2.5751 

3.5081 

4.7649 

6.4534 

32 

33 

1.6345 

1.9222 

.2.2589 

2.6523 

3.6484 

5.0032 

6.8406 

33 

34 

1.6590 

1.9607 

2.3153 

2.7319 

3.7943 

5.2533 

7.2510 

34 

35 

1.6839 

1.9999 

2.3732 

2.8139 

3.9461 

5.5160 

7.6861 

35 

36 

1.7091 

2.0399 

2.4325 

2.8983 

4.1039 

5.7918 

8.1473 

36 

37 

1.7348 

2.0807 

2.4933 

2.9852 

4.2681 

6.0814 

8.6361 

37 

38 

1.7608 

2.1223 

2.5557 

3.0748 

4.4388 

6.3855 

9.1543 

38 

39 

1.7872 

2.1647 

2.6196 

3.1670 

4.6164 

6.7048 

9.7035 

39 

40 

1.8140 

2.2080 

2.6851 

3.2620 

4.8010 

7.0400 

10.2857 

40 

50 

2.1052 

2.6916 

3.4371 

4.3839 

7.1067 

11.4674 

18.4202 

50 

60 

2.4432 

3.2810 

4.3998 

5.8916 

10.5196 

18.6792 

32.9877 

60 

70 

2.8355 

3.9996 

5.6321 

7.9178 

15.5716 

30.4264 

59.0759 

70 

80 

3.2907 

4.8754 

7.2096 

10.6409 

23.0498 

49.6514 

105.7960 

80 

90 

3.8190 

5.9431 

9.2289 

14.3005 

34.1193 

80.7304 

189.4645 

90 

100 

4.4321 

7.2447 

11.8137 

19.2186 

50.5050 

131.5013 

339.3021 

100 

Years. 

li%. 

2%. 

z*%. 

3%. 

4%. 

5%. 

6%. 

Years. 

TABLES 


515 


The  present  value  of  1  due  in  n  years.    vn  —  (1  +  \}~n. 


Years. 

11%. 

2%. 

•1%. 

3%. 

4%. 

*%. 

61  • 

Years. 

1 

0.9852 

0.9804 

0.9750 

0.9709 

0.9615 

0.9524 

0.9434 

1 

2 

0.9707 

0.9012 

0.9518 

0.9426 

0.9246 

0.9070 

0.8900 

2 

3 

0.9563 

0.9423 

0.9286 

0.9151 

0.8890 

0.8638 

0.8396 

3 

4 

0.9422 

0.9238 

0.9060 

0.8885 

0.8548 

0.8227 

0.7921 

4 

5 

0.9283 

0.9057 

0.8839 

0.8626 

0.8219 

0.7835 

0.7473 

5 

6 

0.9145 

0.8880 

0.8623 

0.8375 

0.7903 

0.7462 

0.7050 

6 

7 

0.9010 

0.8706 

0.8413 

0.8131 

0.7599 

0.7107 

0.6651 

1 

8 

0.8877 

0.8535 

0.8207 

0.7894 

0.7307 

0.6768 

0.6274 

8 

9 

0.8746 

0.8368 

0.8007 

0.7664 

0.7026 

0.6446 

0.5919 

9 

10 

0.8617 

0.8203 

0.7812 

0.7441 

0.6756 

0.6139 

0.5584 

10 

11 

0.8489 

0.8043 

0.7621 

0.7224 

0.6496 

0.5847 

0.5268 

11 

12 

0.8364 

0.7885 

0.7436 

0.7014 

0.6246 

0.5568 

0.4970 

12 

13 

0.8240 

0.7730 

0.7254 

0.6810 

0.6006 

0.5303 

0.4688 

13 

14 

0.8118 

0.7579 

0.7077 

0.6611 

0.5775 

0.5051 

0.4423 

14 

15 

0.7999 

0.7430 

0.6905 

0.6419 

0.5553 

0.4810 

0.4173 

15 

10 

0.7880 

0.7284 

0.6736 

0.6232 

0.5339 

0.4581 

0.3936 

16 

17 

0.7764 

0.7142 

0.6572 

0.6050 

0.5134 

0.4363 

0.3714 

17 

18 

0.7649 

0.7002 

0.6412 

0.5874 

0.4936 

0.4155 

0.3503 

18 

It 

0.7536 

0.6864 

0.6255 

0.5703 

0.4746 

0.3957 

0.3305 

19 

'.Ml 

0.7425 

0.6730 

0.6103 

0.5537 

0.4564 

0.3769 

0.3118 

20 

21 

0.7315 

0.6598 

0.5954 

0.5375 

0.4388 

0.3589 

0.2942 

21 

22 

0.7207 

0.6468 

0.5809 

0.5219 

0.4220 

0.3418 

0.2775 

22 

23 

0.7100 

0.6342 

0.5667 

0.5067 

0.4057 

0.3256 

0.2618 

23 

24 

0.6995 

0.6217 

0.5529 

0.4919 

0.3901 

0.3101 

0.2470 

24 

26 

0.6892 

0.6095 

0.5394 

0.4776 

0.3751 

0.2953 

0.2330 

25 

26 

0.6790 

0.5976 

0.5262 

0.4637 

0.3607 

0.2812 

0.2198 

26 

27 

0.6690 

0.5859 

0.6134 

0.4502 

0.3468 

0.2678 

0.2074 

27 

28 

0.6591 

0.5744 

0.5009 

0.4371 

0.3335 

0.2551 

0.1956 

28 

29 

0.6494 

0.5631 

0.4887 

0.4243 

0.3207 

0.2429 

0.1846 

29 

30 

0.6398 

0.5521 

0.4767 

0.4120 

0.3083 

0.2314 

0.1741 

30 

31 

0.6303 

0.5412 

0.4651 

0.4000 

0.2965 

0.2204 

0.1643 

31 

32 

0.6210 

0.5306 

0.4538 

0.3883 

0.2851 

0.2099 

(1.1.  •>.'>(  I 

32 

:n 

0.6118 

0.5202 

0.4427 

0.3770 

0.2741 

0.1999 

0.1462 

33 

34 

0.6028 

0.5100 

0.4316 

0.3660 

0.2636 

0.1904 

0.1379 

34 

35 

0.5939 

0.5000 

n.  tin  t 

0.3554 

0.2534 

0.1813 

0.1301 

35 

M 

0.5851 

0.4902 

0.4111 

0.3450 

0.2437 

0.1727 

0.1227 

36 

37 

0.5764 

0.4806 

0.4011 

0.3350 

0.2343 

0.1644 

0.1158 

37 

38 

0.5679 

0.4712 

0.3913 

0.3252 

0.2253 

0.1566 

0.1092 

38 

39 

0.5595 

0.4620 

0.3817 

0.3158 

0.2166 

0.1491 

0.1031 

39 

40 

0.5513 

0.4529 

0.3724 

0.3066 

0.2083 

0.1420 

0.0972 

40 

50 

0.4750 

0.3715 

0.2909 

0.2281 

0.1407 

0.0872 

0.0543 

50 

60 

0.4093 

0.3048 

0.2273 

0.1697 

0.0951 

0.0535 

0.0303 

60 

70 

0.3627 

0.2500 

0.1776 

0.1263 

0.0642 

0.0329 

0.0169 

70 

80 

0.3039 

0.2061 

0.1387 

0.0940 

0.0434 

0.0202 

0.0095 

80 

90 

0.2619 

0.1683 

0.1084 

0.0699 

0.0293 

0.0124 

0.0053 

90 

100 

0.2256 

0.1380 

0.0846 

0.0520 

0.0198 

0.0076 

0.0029 

100 

Years. 

11%. 

2%. 

21%. 

«%. 

4%. 

«%• 

e%. 

Years. 

516 


UNIFIED  MATHEMATICS 


The  accumulation  of  an  annuity  of  1  per  annum  at  the  end  of  n  years. 

_(l  +  i)»-l 


•sn    i 

Years.  1J%.     2%. 

2}%. 

3%. 

4%.     «%.     «%.  Years. 

1 

1.0000 

1.0000 

1.0000 

1.0000 

1.0000 

1.0000 

1  0000 

1 

2 

2.0150 

2.0200 

2.0250 

2.0300 

2.0400 

2.0500 

2.0600 

2 

3 

3.0452 

3.0604 

3.0756 

3.0909 

3.1216 

3.1525 

3.1836 

3 

4 

4.0909 

4.1216 

4.1525 

4.1836 

4.2465 

4.3101 

4  3746 

4 

5 

5.1523 

5.2040 

5.2563 

5.3091 

5.4163 

5.5256 

5.6371 

5 

6 

6.2296 

6.3081 

6.3877 

6.4684 

6.6330 

6.8019 

6.9753 

6 

7 

7.3230 

7.4343 

7.5474 

7.6625 

7.8983 

8.1420 

8.3938 

7 

8 

8.4328 

8.5830 

8.7361 

8.8923 

9.2142 

9.5491 

9.8975 

8 

9 

9.5593 

9.7546 

9.9545 

10.1591 

10.5828 

11.0266 

11.4913 

t 

10 

10.7027 

10.9497 

11.2034 

11.4638 

12.0061 

12.5779 

13.1808 

10 

11 

11.8633 

12.1687 

12.4835 

12.8078 

13.4864 

14.2068 

14.9716 

11 

12 

13.0412 

13.4121 

13.7956 

14.1920 

15.0258 

15.9171 

16.8699 

12 

13 

14.2368 

14.6803 

15.1404 

15.6178 

16.6268 

17.7130 

18.8821 

13 

14 

15.4504 

15.9739 

16.5190 

17.0863 

18.2919 

19.5986 

21.0151 

14 

15 

16.6821 

17.2934 

17.9319 

18.5989 

20.0236 

21.5786 

23.2760 

15 

16 

17.9324 

18.6393 

19.3802 

20.1569 

21.8245 

23.6575 

25.6725 

16 

17 

19.2014 

20.0121 

20.8647 

21.7616 

23.6975 

25.8404 

28.2129 

17 

18 

20.4894 

21.4123 

22.3863 

23.4144 

25.6454 

28.1324 

30.9057 

18 

19 

21.7967 

22.8406 

23.9460 

25.1169 

27.6712 

30.5390 

33.7600 

19 

20 

23.1237 

24.2974 

25.5447 

26.8704 

29.7781 

33.0660 

36.7856 

20 

21 

24.4705 

25.7833 

27.1833 

28.6765 

31.9692 

35.7193 

39.9927 

21 

ifi, 

25.8376 

27.2990 

28.8629 

30.5368 

34.2480 

38.5052 

43.3923 

22 

23 

27.2251 

28.8450 

30.5844 

32.4529 

36.6179 

41.4305 

46.9958 

23 

24 

28.6335 

30.4219 

32.3490 

34.4265 

39.0826 

44.5020 

50.8156 

24 

25 

30.0630 

32.0303 

34.1578 

36.4593 

41.6459 

47.7271 

54.8645 

25 

26 

31.5140 

33.6709 

36.0117 

38.5530 

44.3117 

51.1135 

59.1564 

26 

27 

32.9867 

35.3443 

37.9120 

40.7096 

47.0842 

54.6691 

63.7058 

27 

28 

34.4815 

37.0512 

39.8598 

42.9309 

49.9676 

58.4026 

68.5281 

28 

29 

35.9987 

38.7922 

41.8563 

45.2189 

52.9663 

62.3227 

73.6398 

29 

30 

37.5387 

40.5681 

43.9027 

47.5754 

56.0849 

66.4389 

79.0582 

30 

31 

39.1018 

42.3794 

46.0003 

50.0027 

59.3283 

70.7608 

84.8017 

31 

32 

40.6883 

44.2270 

48.1503 

52.5028 

62.7015 

75.2988 

90.8898 

32 

33 

42.2986 

46.1116 

50.3540 

55.0778 

66.2095 

80.0638 

97.3432 

33 

34 

43.9331 

48.0338 

52.6129 

57.7302 

69.8579 

85.0670 

104.1838 

34 

35 

45.5921 

49.9945 

54.9282 

60.4620 

73.6522 

90.3203 

111.4348 

35 

36 

47.2760 

51.9944 

57.3014 

63.2759 

77.5983 

95.8363 

119.1209 

36 

37 

48.9851 

54.0343 

59.7339 

66.1742 

81.7022 

101.6281 

127.2681 

37 

38 

50.7199 

56.1149 

62.2273 

69.1594 

85.9703 

107.7095 

135.9042 

38 

39 

52.4807 

58.2372 

64.7830 

72.2342 

90.4092 

114.0950 

145.0585 

39 

40 

54.2679 

60.4020 

67.4026 

75.4013 

95.0255 

120.7998 

154.7620 

40 

50 

73.6828 

84.5794 

97.4843 

112.7969 

152.6671 

209.3480 

290.3359 

50 

60 

96.2147 

114.0515 

135.9916 

163.0534 

237.9907 

353.5837 

533.1282 

60 

70 

122.3638 

149.9779 

185.2841 

230.5941 

364.2905 

588.5285 

967.9322 

70 

80 

152.7109 

193.7720 

248.3827 

321.3630 

551.2450 

971.2288 

1746.5999 

80 

90 

187.9299 

247.1567 

329.1543 

443.3489 

827.9833 

1594.6073 

3141.0752 

90 

100  228.8030 

312.2323 

432.5487 

607.2877 

1237.6237 

2610.0252 

5638.3681 

100 

Years 

•  11%. 

2%. 

**%• 

8% 

4%. 

5%. 

6%.  Years. 

TABLES 

The  present  value  of  an  annuity  of  1  for  n  years, 
1  -  vn 


517 


Years. 

U%. 

2%. 

"i 

r»|       I 

3%. 

4%. 

5%. 

6%. 

Years. 

1 

0.9852 

0.9804' 

0.9756 

0.9709 

0.9615 

0.9524 

0.9434 

1 

2 

1.9559 

1.9416 

1.9274 

1.9135 

1.8861 

1.8594 

1.8334 

2 

3 

2.9122 

2.8839 

2.8560 

2.8286 

2.7751 

2.7232 

2.6730 

3 

4 

3.8544 

3.8077 

3.7620 

3.7171 

3.6299 

3.5460 

3.4651 

4 

5 

4.7827 

4.7135 

4.6458 

4.5797 

4.4518 

4.3295 

4.2124 

5 

6 

5.6972 

5.6014 

5.5081 

5.4172 

5.2421 

5.0757 

4.9173 

6 

7 

6.5082 

6.4720 

6.3494 

6.2303 

6.0021 

5.7864 

5.5824 

7 

8 

7.4859 

7.3255 

7.1701 

7.0197 

6.7327 

6.4632 

6.2098 

8 

g 

8.3808 

8.1622 

7.9709 

7.7861 

7.4353 

7.1078 

6.8017 

9 

10 

9.2222 

8:9826 

8.7521 

8.5302 

8.1109 

7.7217 

7.3601 

10 

11 

10.0711 

9.7868 

9.5142 

9.2526 

8.7605 

8.3064 

7.8869 

11 

12 

10.9075 

10.5753 

10.2578 

9.9540 

9.3851 

8.8633 

8.3838 

12 

13 

11.7315 

11.3484 

10.9832 

10.6350 

9.9856 

9.3936 

8.8527 

13 

14 

12.5434 

12.1062 

11.6909 

11.2961 

10.5631 

9.8986 

9.2950 

14 

15 

13.3432 

12.8493 

12.3814 

11.9379 

11.1184 

10.3797 

9.7122 

15 

16 

14.1313 

13.5777 

13.0550 

12.5611 

11.6523 

10.8378 

10.1059 

16 

17 

14.0076 

14.2919 

13.7122 

13.1661 

12.1657 

11.2741 

10.4773 

17 

18 

15.6726 

14.9920 

14.3534 

13.7535 

12.6593 

11.6896 

10.8276 

18 

19 

16.4262 

15.6786 

14.9789 

14.3238 

13.1340 

12.0853 

11.1581 

19 

20 

17.1686 

16.3514 

15.5892 

14.8775 

13.5903 

12.4622 

11.4699 

20 

21 

17.9001 

17.0112 

16.1845 

15.4150 

14.0292 

12.8212 

11.7641 

21 

22 

18.6208 

17.6580 

16.7654 

15.9369 

14.4511 

13.1630 

12.0416 

22 

23 

19.3309 

18.2922 

17.3321 

16.4436 

14.8568 

13.4886 

12.3034 

23 

24 

20.0301 

18.9139 

17.8850 

16.9355 

15.2470 

13.7986 

12.5504 

24 

25 

20.7196 

19.5235 

18.4244 

17.4131 

15.6221 

14.0940 

12.7834 

25 

26 

21.3986 

20.1210 

18.9.506 

17.8768 

15.9828 

14.3752 

13.0032 

26 

27 

22.0676 

20.7069 

10  4040 

18.3270 

16.3296 

14.6430 

13.2105 

27 

28 

22.7267 

21.2813 

19.9649 

18.7641 

16.6631 

14.8981 

13.4062 

28 

29 

23.3761 

21.8444 

20.4535 

19.1885 

16.9837 

15.1411 

.  13.5907 

29 

30 

24.0158 

22.3965 

20.9303 

19.6004 

17.2920 

15.3725 

13.7648 

30 

31 

24.6461 

22.9377 

21.3954 

20.0004 

17.5885 

15.5928 

13.9291 

31 

32 

25.2671 

23.4683 

21.8492 

20.3888 

17.8736 

15.8027 

14.0840 

32 

33 

25.8790 

23.0886 

22.2919 

20.7638 

18.1476 

16.0025 

14.2302 

33 

34 

26.4817 

24.4986 

22.7238 

21.1318 

18.4112 

16.1929 

14.3681 

34 

35 

27.0756 

24.9986 

23.1452 

21.4872 

18.6646 

16.3742 

14.4982 

35 

36 

27.6607 

25.4888 

23.5563 

21.8323 

18.9083 

16.5469 

14.6210 

36 

37 

28.2371 

25.9695 

23.9573 

22.1672 

19.1426 

16.7113 

14.7368 

37 

38 

28.8051 

26.4406 

24.3486 

22.4925 

19.3679 

16  8679 

14.8460 

38 

39 

•"i  :it>ti'i 

26.9026 

24.7303 

22.8082 

19.5845 

17!0170 

14.9491 

39 

40 

20.0158 

27.8566 

25.1028 

23.1148 

19.7928 

17.1591 

15.0463 

40 

50 

34.9997 

31.4236 

28.3623 

25.7298 

21.4822 

18.2559 

15.7619 

50 

60 

30.3803 

34.7609 

30.9087 

27.6756 

22.6235 

18.0203 

16.1614 

60 

70 

43.1549 

37.4987 

32.8979 

29.1234 

23.3945 

19.3427 

16.3845 

70 

80 

46.4073 

39.7445 

34.4518 

30.2008 

23.91.54 

19.5965 

16.5091 

80 

90 

49.2099 

41.5869 

35.6658 

31.0024 

24.2673 

19.7523 

16.5787 

90 

100 

51.6247 

43.0983 

36.6141 

31.5989 

24.5050 

19.8479 

16.6175 

100 

Years.     1§%. 


21%. 


3%. 


4%. 


5%. 


6'  ,  .    Years. 


518 


UNIFIED  MATHEMATICS 


The  annual  sinking  fund  which  will  accumulate  to  1  at  the  end  of  n  years. 


1 


s> 

wl   C1  " 

M)"- 

—  -    JL  w  * 

,/  k?  IC&JLU 

a 

O.U.U.  C. 

5] 

»M*W 

<S'» 

Years. 

11%. 

2%. 

*J%. 

3%. 

4%. 

5%. 

6%. 

Years. 

1 

1.0000 

1.0000 

1.0000 

1.0000 

1.0000 

1.0000 

1.0000 

1 

2 

-0.4963 

0.4950 

0.4938 

0.4926 

0.4902 

0.4878 

0.4854 

2 

3 

0.3284 

0.3268 

0.3251 

0.3235 

0.3203 

0.3172 

0.3141 

3 

4 

0.2444 

0.2426 

0.2408 

0.2390 

0.2355 

0.2320 

0.2286 

4 

5 

0.1941 

0.1922 

0.1902 

0.1884 

0.1846 

0.1810 

0.1774 

5 

6 

0.1605 

0.1585 

0.1566 

0.1546 

0.1508 

0.1470 

0.1434 

6 

7 

0.1366 

0.1345 

0.1325 

0.1305 

0.1266 

0.1228 

0.1191 

7 

8 

0.1186 

0.1165 

0.1145 

0.1125 

0.1085 

0.1047 

0.1010 

8 

9 

0.1046 

0.1025 

0.1005 

0.0984 

0.0945 

0.0907 

0.0870 

9 

10 

0.0934 

0.0913 

0.0893 

0.0872 

0.0833 

0.0795 

0.0759 

10 

11 

0.0843 

0.0822 

0.0801 

0.0781 

0.0741 

0.0704 

0.0668 

11 

12 

0.0767 

0.0746 

0.0725 

0.0705 

0.0666 

0.0628 

0.0593 

12 

13 

0.0702 

0.0681 

0.0660 

0.0640 

0.0601 

0.0565 

0.0530 

13 

14 

0.0647 

0.0626 

0.0605 

0.0585 

0.0547 

0.0510 

0.0476 

14 

15 

0.0599 

0.0578 

0.0558 

0.0538 

0.0499 

0.0463 

0.0430 

15 

16 

0.0558 

0.0537 

0.0516 

0.0496 

0.0458 

0.0423 

0.0390 

16 

17 

0.0521 

0.0500 

0.0479 

0.0460 

0.0422 

0.0387 

0.0354 

17 

18 

0.0488 

0.0467 

0.0447 

0.0427 

0.0390 

0.0355 

0.0324 

18 

19 

0.0459 

0.0438 

0.0418 

0.0398 

0.0361 

0.0327 

0.0296 

19 

20 

0.0432 

0.0412 

0.0391 

0.0372 

0.0336 

0.0302 

0.0272 

20 

21 

0.0409 

0.0388 

0.0368 

0.0349 

0.0313 

0.0280 

0.0250 

21 

22 

0.0387 

0.0366 

0.0346 

0.0327 

0.0292 

0.0260 

0.0230 

22 

23 

0.0367 

0.0347 

0.0327 

0.0308 

0.0273 

0.0241 

0.0213 

23 

24 

0.0349 

0.0329 

0.0309 

0.0290 

0.0256 

0.0225 

0.0197 

24 

25 

0.0333 

0.0312 

0.0293 

0.0274 

0.0240 

0.0210 

0.0182 

25 

26 

0.0317 

0.0297 

0.0278 

0.0259 

0.0226 

0.0196 

0.0169 

26 

27 

0.0303 

0.0283 

0.0264 

0.0246 

0.0212 

0.0183 

0.0157 

27 

28 

0.0290 

0.0270 

0.0251 

0.0233 

0.0200 

0.0171 

0.0146 

28 

29 

0.0278. 

0.0258 

0.0239 

0.0221 

0.0189 

0.0160 

0.0136 

29 

30 

0.0266 

0.0246 

0.0228 

0.0210 

0.0178 

0.0151 

0.0126 

30 

31 

0.0256 

0.0236 

0.0217 

0.0200 

0.0169 

0.0141 

0.0118 

31 

32 

0.0246 

0.0226 

0.0208 

0.0190 

0.0159 

0.0133 

0.0110 

32 

33 

0.0236 

0.0217 

0.0199 

010182 

0.0151 

0.0125 

0.0103 

33 

34 

0.0228 

0.0208 

0.0190 

0.0173 

0.0143 

0.0118 

0.0096 

34 

35 

0.0219 

0.0200 

0.0182 

0.0165 

0.0136 

0.0111 

0.0090 

35 

36 

0.0212 

0.0192 

0.0175 

0.0158 

0.0129 

0.0104 

0.0084 

36 

37 

0.0204 

0.0185 

0.0167 

0.0151 

0.0122 

0.0098 

0.0079 

37 

38 

0.0197 

0.0178 

0.0161 

0.0145 

0.0116 

0.0093 

0.0074 

38 

39 

0.0191 

0.0172 

0.0154 

0.0138 

0.0111 

0.0088 

0.0069 

39 

40 

0.0184 

0.0166 

0.0148 

0.0133 

0.0105 

0.0083 

0.0065 

40 

50 

0.0136 

0.0118 

0.0103 

0.0089 

0.0066 

0.0048 

0.0034 

50 

60 

0.0104 

0.0088 

0.0074 

0.0061 

0.0042 

0.0028 

0.0019 

60 

70 

0.0182 

0.0067 

0.0054 

0.0043 

0.0027 

0.0017 

0.0010 

70 

80 

0.0065 

0.0052 

0.0040 

0.0031 

0.0018 

0.0010 

0.0006 

80 

90 

0.0053 

0.0040 

0.0030 

0.0023 

0.00121 

0.00063 

0.00032 

90 

100 

0.0044 

0.0032 

0.0023 

0.0016 

0.00081 

0.00038 

0.00018 

100 

Years.     1J%.         2%.          2J%. 


3%. 


4%. 


6%.       Years. 


INDEX 


Abel,  204,  399. 
abscissa,  67. 
absolute  value,  104. 
addition,  formulas,  237. 

geometrical,  13. 

of  numbers,  4. 
aeroplane,  277,  278,  388. 
Ahmes  papyrus,  90. 
air-pump,  179  ff. 
Al-Battani,  126,  216. 
Alexander  III  Bridge  in  Paris,  309. 
algebra,  fundamental  theorem,  392. 

literal,  12. 

algebraic  functions,  66  ff.,  392  ff. 
Almagest,  125. 
amplitude,  of  complex  number,  444. 

of  S.  H.  M.,  412. 

of  sinusoid,  408,  410. 
angle,  between  two  lines,  246  ff.,  461. 

depression,  210. 
'    direction,  457. 

measurement,  110  ff. 

of  incidence,  262  ff. 

of  refraction,  262  ff. 
annuity,  183  ff. 

approximations,    logarithms,  48,   140 
ff.,  274. 

numerical,  31  ff.,  93. 

trigonometric,  170  ff. 
anti-sine,  -cosine,  -tangent,  etc.,  137. 
Arabic    mathematics,    90,    119,    126, 

203,  256. 

Archimedes,  183,  404,  438. 
arc  sin,  -cos,  -tan,  etc.,  137. 
area,  of  an  ellipse,  287  ff. 

of  an  inscribed  quadrilateral,  261. 

of  a  triangle,  51,  209  ff.,  258  ff.,  261. 
Argand,  450. 
arithmetical  series,  166  ff. 

mean,  172  ff. 
asymptote,  325. 
auxiliary  circles,  282  ff. 


Babylonian  mathematics,  111,  183. 

bacteria,  growth,  423,  427. 

barometric  pressure,  60,  426  ff. 

Bhaskara,  261. 

binomial  series,  193  ff. 

biquadratic,  392,  399. 

bisector,  of    the  angle  between  two 

lines,  160  ff. 
perpendicular,  151,  164. 
Briggs,  49. 

capacity,  of  a  can,  53,  73,  83,  100. 

of  a  cistern,  94. 
Cardan,  399. 
cardioid,  437. 
Carrel,  430. 

centigrade  scale,  2,  86,  103. 
centimeters  and  inches,  103. 
characteristic,  44. 
chess-board  problem,  187. 
Chinese  mathematics,  203. 
circle,  220  ff. 

auxiliary,  282  ff. 

circular  sections  of  a  cone,  etc.,  489  ff. 
circumferential  velocity,  115. 
cissoid,  437. 

Colosseum  in  Rome,  288,  362. 
compass,  geometrical,  4. 

mariner's,  114. 
complex  numbers,  439  ff. 
components  of  a  vector,  152  ff. 
compound  interest,  54  ff. 
conchoid  of  Nicomedes,  437. 
congruent  angles,  116. 
conjugate  hyperbolas,  328. 

numbers,  441. 
cone,  483. 

conic  sections,  287,  483  ff. 
connecting  rod,  421  ff. 
continuous  functions,  393,  397. 
coordinate  axes,  453. 

planes,  453. 


519 


520 


INDEX 


coordinates,  67  ff.,  115  ff.,  118,  435, 

453. 

cosecant,  definition,  120. 
cosine,  definition,  117  ff. 

law,  252  ff. 

cotangent,  definition,  120,  125  ff. 
crank  arm,  421  ff. 
cubic,  392,  399. 

curves,  466. 
cubical  parabola,  74. 
curvature  of  the  earth,  146. 
cyclic  interchange,  252. 
cylindrical  surfaces,  466  ff. 

damped  vibrations,  431  ff. 
decagon,  124. 
decimal,  recurring,  177. 
deflection  angle,  217. 
De  Moivre,  445. 
departure,  210. 
depression,  210. 
Descartes,  69,  449. 
dihedral  angle,  456. 
dip,  210. 
directed  line,  1. 
direction  angles,  457,  470. 

cosines,  457,  462. 
directrix,  of  a  conic,  289,  309,  320. 

of  a  cylinder,  467. 
discount,  190,  386  ff. 
discriminant,  89. 

distance,  between  two  points,  104  ff., 
458. 

from  a  point  to  a  line,  158  ff.,  487. 

from  a  point  to  a  plane,  471. 
division,  5. 

abbreviated,  35. 

graphical,  9,  10. 

synthetic,  25,  394  ff. 
duplication  of  a  cube,  204  ff. 

eccentricity  (e),  289,  297,  347. 
Egyptian  mathematics,  90,  175,  183. 
electrical      phenomena,      108,     412, 

415  ff.,  432. 

element  of  a  cylinder,  467. 
ellipse,  280  ff. 
ellipsoids,  476  ff. 
elliptical  arch,  295,  307,  355. 

gears,  356  ff. 
elliptic  paraboloid,  481  ff. 
equilateral  hyperbola,  327,  329. 
Euclid,  183. 


Euler,  243. 
evolution,  41. 
explicit  function,  58. 
exponent,  40. 

Fahrenheit  scale,  2,  86,  103. 
family  of  surfaces,  465  ff. 
Farm  Loan  Act,  188  ff. 
Fermat.  69. 
Ferrari,  399. 
Fieri,  399. 
fly-wheel,  434. 
focal  distances,  291,  327. 
focus,  289,  299  ff.,  305,  309. 
fourth  dimension,  452  ff. 
Franklin,  56. 
frequency,  412. 
functions,  58, 

algebraic,  66  ff. 

linear,  77  ff.,  101  ff.,  149  ff.,  155  ff., 
462  ff. 

quadratic,  87  ff. 

trigonometric,  110  ff. 

Galton,  179. 

Gauss,  392,  450. 

generator,  of  a  cylinder,  467. 

of  a  surface,  491. 
geometrical  mean,  178. 

series,  176  ff. 
Glover,  188. 

graphical  methods,  13ff.,  169ff.,180ff. 
greater,  1,  3. 

Greek  mathematics,    119,    125,    183, 
242,  256  ff.,  287. 

half-angle  formulas,  trigonometric 
functions,  247  ff. 

oblique  triangle,  258. 
Halley,  428. 

healing  of  a  wound,  law,  429  ff. 
helical  spring,  414  ff. 
Hero,  51,  69,  261. 
Hill  Auditorium,  352,  362,  389. 
Hindu  mathematics,  119,  261. 
hyperbola,  320  ff. 
hyperbolic  paraboloid,  481  ff. 
hyperboloid,  of  one  sheet,  475,  479  ff. 

of  two  sheets,  475,  479  ff. 

imaginary  numbers,  439  ff.,  449,  450. 
implicit  function,  58. 
index  of  refraction,  262  ff. 
indices,  theory  of,  40  ff. 


INDEX 


521 


induction,  mathematical.  167,  198. 
infinity,  97  ff.,  176  ff.,   181,  200   S. 

311  ff.,  323  ff. 
integers,  3. 
intercepts,  82  ff. 

interest,  54  ff.,  92,  183  ff.,  200,  423. 
interpolation,  46  ff.,  140  ff.,  170  ff. 
intersections  of  graphs,  80  ff.,  465. 
inverse  functions,  137. 
involution,  41. 
irrational  numbers,  10  ff. 

Kepler,  2X7. 
Khowarizmi,  90. 

latitude,  67,  147. 

lead  of  a  screw,  211. 

Leibniz,  69. 

light-waves,  415. 

limagon.  437. 

limiting  values,  97  ff. 

linear  function,  77  ff.,  101  ff.,  149  ff., 

.   155  ff.,  462  ff. 
logarithms,  41  ff. 
London  Bridge,  281,  284. 
longitude,  67,  147. 

mantissa,  44. 

maximum,  398. 

mean,  arithmetical,  172  ff. 

geometrical,  178. 

weighted,  173. 
measurements,  31  ff.,  37. 
mercury,  expansion  of,  63,  85,  103. 
mid-point  formula,  107. 
minimum,  398. 
minutes.  111. 
modulus,  444  ff. 
Mohammed  ibn  Musa  al-Khowarizmi, 

90. 
multiplication,  5. 

abbreviated,  34. 

graphical,  15. 
musical  scale,  414. 

Napier,  49. 

Nasir  ad-Din  at-Tusi,  256. 
natural  logarithms,  425  ff. 
negative  angles,  110. 

numbers,  6. 

Newton,  69,  203  ff.,  260  ff.,  287. 
Nicomedes,  437. 


normal  form,  of   the  equation  of  a 
line,  155  ff. 

of  a  plane,  470  ff. 
normal  distribution  curve,  428. 
du  Notiy,  431. 
numbers,  1  ff. 

classification,  3. 

definition,  3. 

oblate  spheroid,  476,  478. 
Oldenburg,  203. 
Omar  al-Khayyam,  204. 
one-to-one  correspondence,  1. 
orbit  of  the  earth,  38,  297  ff. 
ordinate,  67. 
organ-pipes,  388. 
origin,  1,  453. 

parabola,  73,  309  ff. 
parabolic  arch,  317. 

reflector,  318,  319. 
paraboloids,  481  ff. 
parallel  lines,  149,  462. 

planes,  471. 
parameter,  109,  221,  281,  326,  457  ff., 

464,  468. 
Pascal,  203,  437. 
pendulum,  38,  39,  54, 100,  213,  317  ff., 

427,  434. 

percentage  error,  31,  36. 
perpendicular  lines,  149,  462. 
phase-angle,  412,  416. 
piston-rod  motion,  420  ff. 
point  of  division,  105  ff.,  108,  459. 
point-slope  formula,  101. 
polar-coordinates,  115  ff.,  118,  435  ff. 
population  statistics,  57,  65  ff.,  179. 
premium,  190. 
present  value,  184  ff. 
progressive  computation,  195,  201  ff. 
projectiles,  87,  93,  94,  99,  100,  154, 

214  ff.,  278  ff.,  317. 
projecting  planes,  468  ff. 
projection  of  vectors,  152  ff.,  156  ff. 
prolate  spheroid,  476,  478. 
Ptolemy,  125,  242. 
Pythagoras,  125. 

quadrants,  113. 
quadratic  form,  95  ff. 

function,  87  ff. 

graphical  solution,  90  ff. 

solution,  88  ff. 


522 


INDEX 


radian,  111  ff. 
radical  axis,  231  ff. 

center,  233. 
railroad  curves,  217  ff. 
rational  numbers,  7. 
real  numbers,  2. 
Recorde,  13. 

rectangular  hyperbola,  327,  329. 
rectilinear  generator,  491  ff. 
reflection  of  light,  261  ff. 
refraction  of  light,  261  ff. 
Regiomontanus,  257. 
regular  polygons,  448,  450. 
related  angles,  128  ff. 
remainder  theorem,  26,  394  ff. 
resultant,  152. 

Rialto  in  Venice,  213,  225,  236. 
right  focal  chord,  292,  310,  323. 
right-handed  system  of  axes,  454. 
roots  of  unity,  447  ff. 
rotation,  positive  and  negative,  110. 
ruled  surfaces,  491  ff. 

scalar  line,  1. 

screw,  211. 

secant,  definition,  120. 

seconds,  211. 

series,  arithmetical,  166  ff. 

binomial,  193  ff. 

geometrical,  176  ff. 
significant  figures,  31. 
silo,  53,  76,  100. 
simple  curve,  217. 
simple  harmonic  motion,  409,  412. 
sine  curve,  407  ff. 

definition,  117. 

law,  255  ff. 
sinking  fund,  189  ff. 
sinusoid,  407  ff.,  419. 
skew  lines,  461. 
slope,  101,  149  ff.,  397  ff. 
slope-intercept  formula,  101. 
sound,  86,  99,  107,  306,  412  ff. 
specific  gravity,  39. 
sphere,  458,  473,  476. 
spherical  segment,  405. 
spheroids,  478. 


spiral,  211,  438. 
squaring  numbers,  21,  73. 
statistics,  59  ff. 
Steinmetz,  450. 
subtraction,  4. 

geometrical,  13. 
symbols,  13. 
synthetic  division,  25,  394  ff. 

tables,  46  ff.,  139  ff.,  495  ff. 
tabular  difference,  48,  141. 
tangent,  definition,  119,  125  ff. 

law,  261. 

planes,  490  ff. 

to  a  curve,  227,  230,  298  ff.,  301  ff., 

315  ff.,  490  ff. 
Tartaglia,  399. 
temperature  chart,  60. 
tensor,  444. 
Thales,  266. 

transformation  of  coordinates,  371  ff. 
transition  curves,  217,  389  ff. 
triangle,  trigonometric  solution,  251 

ff.,  266  ff. 

trisection  of  an  angle,  399. 
two-point  formula  of  a  line,  101. 
tuning  fork,  412. 

variable,  12. 

variation  of  trigonometric  functions, 

126  ff. 
vector,  115,  152  ff.,  444  ff. 

in  space,  456. 
vectorial  angle,  115. 
vibrations,  407,  412  ff. 
Viete,  13,  242. 
voice  records,  413. 

water,  weight,  volume,  etc.,  61  ff., 

85,  103. 

wave  lengths,  413. 
wave  motion,  407. 
Wessel,  449. 
Widmann,  13. 

Zeno,  177. 


v 


DC  SOUTHERN  REGIONAL  LIBRARY  FAOUTY 


A    000932101     9 


'    J0U 


tOBA»GBLB8> 


